13 electrons have spin in one direction and 16 electrons in other direction

All the electrons have spin in one direction

$4{{\rm{K}}_2}{\rm{C}}{{\rm{r}}_2}{{\rm{O}}_7}\xrightarrow{{{\rm{Heat}}}}4{{\rm{K}}_2}{\rm{Cr}}{{\rm{O}}_4} + 3{{\rm{O}}_2} + X$ In the above reaction, $X$ is

${\rm{C}}{{\rm{r}}_2}{{\rm{O}}_3}$

${\rm{Cr}}{{\rm{O}}_3}{\rm{\;}}$

${\rm{C}}{{\rm{r}}_2}{{\rm{O}}_7}$

d & f- BLOCK ELEMENTS MCQs for NEET UG CHEMISTRY — Easy, Page 3

61 Easy

61 of 80Question permalink

Application +4 | -1

Which of the following ions is diamagnetic in nature?

View answer and explanation

Correct option: D

Both (2) and (3)

Option D (Both (2) and (3)) is correct because both La3+ and Zn2+ have completely filled electronic configurations, making them diamagnetic.

Solution approach

Correct Answer + Core Reason: Both (2) and (3). Diamagnetic substances have all their electrons paired.

Step-by-Step Solution:

Step 1: Determine the electronic configurations of each ion.

Step 2: $\mathrm{Cu}^{2+}$ has a $d^9$ configuration, meaning it has one unpaired electron, making it paramagnetic.

Step 3: $\mathrm{La}^{3+}$ has a configuration of $[\mathrm{Xe}]$ , meaning it has no unpaired electrons, making it diamagnetic.

Step 4: $\mathrm{Zn}^{2+}$ has a $d^{10}$ configuration, meaning all its d-orbitals are filled, and it has no unpaired electrons, making it diamagnetic.

Step 5: Therefore, $\mathrm{La}^{3+}$ and $\mathrm{Zn}^{2+}$ are diamagnetic.

Why other options are wrong:
- Option A: $\mathrm{Cu}^{2+}$ is paramagnetic because it has an unpaired electron in its d-orbital.
- Option B: $\mathrm{La}^{3+}$ is diamagnetic, but this option doesn't include $\mathrm{Zn}^{2+}$ , which is also diamagnetic.
- Option C: $\mathrm{Zn}^{2+}$ is diamagnetic, but this option doesn't include $\mathrm{La}^{3+}$ , which is also diamagnetic.

Key Concepts:
- Diamagnetism: All electrons are paired.
- Paramagnetism: Presence of unpaired electrons.
- Electronic configuration of ions.

Memory Aids:
- Diamagnetic: \

62 Easy

62 of 80Question permalink

Application +4 | -1

The magnetic moment (in BM) of

\mathrm{Z}{\mathrm{n}^{2 + }}

ion according to spin-only formula is

View answer and explanation

Correct option: A

Zero

The magnetic moment of

\mathrm{Z}{\mathrm{n}^{2 + }}

ion is zero because

\mathrm{Z}{\mathrm{n}^{2 + }}

has a

3d^{10}

configuration, meaning all d-orbitals are filled and there are no unpaired electrons.

Solution approach

Correct Answer + Core Reason: Zero. $\mathrm{Z}{\mathrm{n}^{2 + }}$ has a $3d^{10}$ configuration with no unpaired electrons.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of Zn.\nZn (atomic number 30): $[Ar] 3d^{10} 4s^2$

Step 2: Determine the electronic configuration of $\mathrm{Z}{\mathrm{n}^{2 + }}$ .
$\mathrm{Z}{\mathrm{n}^{2 + }}$ : $[Ar] 3d^{10}$

Step 3: Calculate the number of unpaired electrons (n).\nSince the $3d$ orbitals are completely filled, there are no unpaired electrons. Therefore, n = 0.

Step 4: Calculate the magnetic moment using the spin-only formula:
$\mu = \sqrt{n(n+2)}$ BM
$\mu = \sqrt{0(0+2)}$ BM
$\mu = 0$ BM

Why other options are wrong:
- Option B: $1.73$ . This would correspond to one unpaired electron, which is not the case for $\mathrm{Z}{\mathrm{n}^{2 + }}$ .
- Option C: $2.84$ . This would correspond to two unpaired electrons, which is not the case for $\mathrm{Z}{\mathrm{n}^{2 + }}$ .
- Option D: $3.87$ . This would correspond to three unpaired electrons, which is not the case for $\mathrm{Z}{\mathrm{n}^{2 + }}$ .

Key Concepts:
- Electronic configuration
- Magnetic moment
- Spin-only formula

Memory Aids:
- Remember that filled or half-filled d-orbitals contribute to stability and affect magnetic properties.

Quick Tips:
- If an ion has a $d^{0}$ or $d^{10}$ configuration, its magnetic moment is zero.

63 Easy

63 of 80Question permalink

Understanding +4 | -1

Choose the correct statement.

View answer and explanation

Correct option: C

Transition elements exhibit variable oxidation states.

Transition elements exhibit variable oxidation states because of the involvement of (n-1)d and ns electrons in bonding.

Solution approach

Correct Answer + Core Reason: Transition elements exhibit variable oxidation states. This is due to the involvement of (n-1)d and ns electrons.

Step-by-Step Solution:

Step 1: Understand the properties of transition elements.

Step 2: Transition elements exhibit variable oxidation states because the energy difference between the (n-1)d and ns orbitals is small, allowing both sets of electrons to participate in bonding.

Step 3: Therefore, the correct statement is that transition elements exhibit variable oxidation states.

Why other options are wrong:
- Option A: Transition elements generally have high melting points due to strong metallic bonding.
- Option B: Transition elements and their compounds often have catalytic activity due to their variable oxidation states and ability to form complexes.
- Option D: Transition elements do not show inert pair effect; this effect is more prominent in p-block elements.

Key Concepts:
- Properties of transition elements
- Variable oxidation states
- Catalytic activity

Memory Aids:
- Transition elements exhibit variable oxidation states.

Quick Tips:
- Variable oxidation states are a key characteristic of transition elements.

64 Easy

64 of 80Question permalink

Remembering +4 | -1

The most stable oxidation state of lanthanides is

View answer and explanation

Correct option: D

+ 3

Option D (

+ 3

) is correct because the +3 oxidation state is the most stable and common oxidation state for lanthanides due to the electronic configuration they attain.

Solution approach

Correct Answer + Core Reason: $+ 3$ . The +3 oxidation state is the most stable for lanthanides.

Step-by-Step Solution:

Step 1: Recall the common oxidation states of lanthanides.

Step 2: The most stable oxidation state for lanthanides is +3 because, after losing three electrons, they attain a stable electronic configuration (either $f^0$ , $f^7$ , or $f^{14}$ ).

Why other options are wrong:
- **Option A ( $+ 2$ ): +2 oxidation state is less common and less stable.
- Option B ( $+ 4$ ): +4 oxidation state is observed in some lanthanides (e.g., Ce), but it is less common than +3.
- Option C (0): The 0 oxidation state refers to the elemental form, not an ion.

Key Concepts:
- Oxidation states of lanthanides
- Stability of electronic configurations

Memory Aids:
- Lanthanides commonly exist as +3 ions.

Quick Tips:**
- Remember that +3 is the most stable oxidation state for lanthanides.

65 Easy

65 of 80Question permalink

Understanding +4 | -1

Transition metals in their compounds show:

View answer and explanation

Correct option: C

Ionic and covalent bonds

Transition metals form both ionic and covalent bonds because they have variable oxidation states and relatively high electronegativity.

Solution approach

Correct Answer + Core Reason: Ionic and covalent bonds. Transition metals exhibit both types of bonding.

Step-by-Step Solution:

Step 1: Understand the bonding characteristics of transition metals.

Step 2: Transition metals can form ionic bonds due to their ability to form cations with varying charges. They can also form covalent bonds due to their relatively high electronegativity and the involvement of d-electrons in bonding.

Step 3: Therefore, transition metals in their compounds show both ionic and covalent bonds.

Why other options are wrong:
- Option A: Ionic bonds alone are not sufficient to describe the bonding in transition metal compounds because covalent character is often present.
- Option B: Covalent bonds alone are not sufficient to describe the bonding in transition metal compounds because ionic character is often present.
- Option D: Ionic and coordinate bonds can be present, but the most general answer includes both ionic and covalent bonds.

Key Concepts:
- Bonding in transition metal compounds
- Ionic and covalent character
- Variable oxidation states

Memory Aids:
- Transition metals can form both ionic and covalent bonds.

Quick Tips:
- Consider both ionic and covalent character when describing the bonding in transition metal compounds.

66 Easy

66 of 80Question permalink

Application +4 | -1

In Cu (Z = 29):

View answer and explanation

Correct option: B

14 electrons have spin in one direction and 15 electrons in other direction

Option B is correct because copper (Z=29) has 15 electrons with one spin and 14 with the opposite spin, considering its electronic configuration [Ar]3d¹⁰ 4s¹.

Solution approach

Correct Answer + Core Reason: 14 electrons have spin in one direction and 15 electrons in other direction. This is based on Hund's rule and filling of orbitals.

Step-by-Step Solution:

Step 1: Write the electronic configuration of copper (Cu). $\nCu (Z = 29): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d¹⁰ or [Ar] 3d¹⁰ 4s¹$

Step 2: Determine the number of electrons with each spin in each orbital.
- 1s²: 1 electron with spin +1/2 and 1 electron with spin -1/2
- 2s²: 1 electron with spin +1/2 and 1 electron with spin -1/2
- 2p⁶: 3 electrons with spin +1/2 and 3 electrons with spin -1/2
- 3s²: 1 electron with spin +1/2 and 1 electron with spin -1/2
- 3p⁶: 3 electrons with spin +1/2 and 3 electrons with spin -1/2
- 4s¹: 1 electron with spin +1/2 (or -1/2, it doesn't matter as long as we are consistent)
- 3d¹⁰: 5 electrons with spin +1/2 and 5 electrons with spin -1/2

Step 3: Sum the number of electrons with each spin.\nAssume 4s¹ has spin +1/2.
- Electrons with spin +1/2: 1 + 1 + 3 + 1 + 3 + 1 + 5 = 15
- Electrons with spin -1/2: 1 + 1 + 3 + 1 + 3 + 5 = 14

Step 4: Conclude the answer.\nTherefore, 15 electrons have spin in one direction, and 14 electrons have spin in the other direction.

Why other options are wrong:
- Option A: 13 electrons have spin in one direction and 16 electrons in other direction: This is incorrect based on the electronic configuration and the filling of orbitals.
- Option C: All the electrons have spin in one direction: This violates the Pauli exclusion principle, which states that no two electrons in an atom can have the same set of quantum numbers.
- Option D: None of the above: This is incorrect because option B is correct.

Key Concepts:
- Electronic configuration
- Hund's rule
- Pauli exclusion principle
- Spin quantum number

Memory Aids:
- Remember Hund's rule: Electrons fill orbitals and subshells individually before pairing up.

Quick Tips:
- Write the electronic configuration of the atom.
- Fill the orbitals according to Hund's rule and the Pauli exclusion principle.
- Count the number of electrons with each spin.

67 Easy

67 of 80Question permalink

Understanding +4 | -1

Most basic compound among the following is

View answer and explanation

Correct option: A

La{\left( {OH} \right)_3}

Option A (

La{\left( {OH} \right)_3}

) is the most basic because the basicity of lanthanide hydroxides decreases as the ionic size decreases from La to Lu.

Solution approach

Correct Answer + Core Reason: $La{\left( {OH} \right)_3}$ . Basicity decreases down the lanthanide series due to decreasing ionic size.

Step-by-Step Solution:

Step 1: Understand the trend of basicity in lanthanide hydroxides.

Step 2: The basicity of lanthanide hydroxides decreases as the ionic size decreases from La to Lu. This is due to the increasing covalent character of the M-OH bond as the size of the lanthanide ion decreases.

Step 3: $La^{3+}$ has the largest ionic radius among the given lanthanides, making $La{\left( {OH} \right)_3}$ the most ionic and hence the most basic.

Step 4: Therefore, $La{\left( {OH} \right)_3}$ is the most basic compound.

Why other options are wrong:
- Option B: $Eu{\left( {OH} \right)_3}$ is less basic than $La{\left( {OH} \right)_3}$ because $Eu^{3+}$ has a smaller ionic radius than $La^{3+}$ .
- Option C: $Er{\left( {OH} \right)_3}$ is less basic than $La{\left( {OH} \right)_3}$ because $Er^{3+}$ has a smaller ionic radius than $La^{3+}$ .
- Option D: $Lu{\left( {OH} \right)_3}$ is the least basic among the given options because $Lu^{3+}$ has the smallest ionic radius.

Key Concepts:
- Basicity of lanthanide hydroxides
- Ionic size and its effect on basicity

Memory Aids:
- Remember that basicity decreases as you move from La to Lu in lanthanide hydroxides.

Quick Tips:
- The larger the ionic radius of the lanthanide ion, the more basic its hydroxide.

68 Easy

68 of 80Question permalink

Application +4 | -1

The +3 ion of which one of the following has half-filled

4f

sunshell?

View answer and explanation

Correct option: C

Gd

Option C is correct because Gd3+ has 7 electrons in its 4f subshell, resulting in a half-filled configuration (4f⁷).

Solution approach

Correct Answer + Core Reason: Gd. Gd³⁺ has 7 electrons in its 4f subshell, which is a half-filled configuration.

Step-by-Step Solution:

Step 1: Write the electronic configuration of the neutral atoms.
- La (Z = 57): [Xe]5d¹6s²
- Lu (Z = 71): [Xe]4f¹⁴5d¹6s²
- Gd (Z = 64): [Xe]4f⁷5d¹6s²
- Ce (Z = 58): [Xe]4f¹5d¹6s²

Step 2: Determine the electronic configuration of the +3 ions.
- La³⁺: [Xe] (loses 5d¹ and 6s² electrons)
- Lu³⁺: [Xe]4f¹⁴ (loses 5d¹ and 6s² electrons)
- Gd³⁺: [Xe]4f⁷ (loses 5d¹ and 6s² electrons)
- Ce³⁺: [Xe]4f¹ (loses 5d¹ and 6s² electrons)

Step 3: Identify the ion with a half-filled 4f subshell.\nA half-filled 4f subshell has 7 electrons (4f⁷). Gd³⁺ has the [Xe]4f⁷ configuration.

Step 4: Conclude the answer.\nTherefore, the correct answer is Gd.

Why other options are wrong:
- Option A: La³⁺ has an electronic configuration of [Xe], which has no 4f electrons.
- Option B: Lu³⁺ has an electronic configuration of [Xe]4f¹⁴, which is a fully filled 4f subshell.
- Option D: Ce³⁺ has an electronic configuration of [Xe]4f¹, which has only one 4f electron.

Key Concepts:
- Electronic configuration
- Lanthanide ions
- Half-filled subshells

Memory Aids:
- Remember that half-filled and fully filled subshells are particularly stable.

Quick Tips:
- Write the electronic configuration of the neutral atom first.
- Remove electrons from the outermost shell when forming cations.

69 Easy

69 of 80Question permalink

Remembering +4 | -1

Most common oxidation state of vanadium is

View answer and explanation

Correct option: D

+ 5

Option D (

+ 5

) is correct because the most stable and common oxidation state of Vanadium is +5, found in compounds like

V_2O_5

.

Solution approach

Correct Answer + Core Reason: $+ 5$ . The most common oxidation state of vanadium is +5.

Step-by-Step Solution:

Step 1: Consider the electronic configuration of Vanadium (V): [Ar] $3d^3 4s^2$ .

Step 2: Vanadium can exhibit various oxidation states, including +2, +3, +4, and +5.

Step 3: The most stable and common oxidation state of Vanadium is +5. This is because it can lose all 5 valence electrons ( $3d^3 4s^2$ ) to achieve a stable electronic configuration.

Step 4: Vanadium(V) oxide ( $V_2O_5$ ) is a common and stable compound of vanadium.

Why other options are wrong:
- Option A: +2 is a possible oxidation state, but not the most common.
- Option B: +3 is a possible oxidation state, but not the most common.
- Option C: +4 is a possible oxidation state, but not the most common.

Key Concepts:
- Oxidation states
- Electronic configuration
- Stability of compounds

Memory Aids:
- Remember that Vanadium can exhibit multiple oxidation states, but +5 is the most common.

Quick Tips:
- Consider the electronic configuration to determine the possible oxidation states.

70 Easy

70 of 80Question permalink

Analyzing +4 | -1

Which of the following compounds has colour but no unpaired electrons?

View answer and explanation

Correct option: A

\mathrm{KMnO}_4

Option A (

KMnO_4

) is correct because it exhibits color due to charge transfer transitions, even though it has no unpaired d-electrons in the traditional sense.

Solution approach

Correct Answer + Core Reason: $\mathrm{KMnO}_4$ . It has colour due to charge transfer transitions despite having no unpaired d-electrons.

Step-by-Step Solution:

Step 1: Determine the oxidation state of Mn in each compound.
$\mathrm{KMnO}_4$ : Mn is in +7 oxidation state.
$\mathrm{K}_2\mathrm{MnO}_4$ : Mn is in +6 oxidation state.
$\mathrm{MnSO}_4$ : Mn is in +2 oxidation state.
$\mathrm{MnCl}_2$ : Mn is in +2 oxidation state.

Step 2: Determine the electronic configuration of Mn in each compound.
$Mn^{7+}$ : Electronic configuration is [Ar] $3d^0$
$Mn^{6+}$ : Electronic configuration is [Ar] $3d^1$
$Mn^{2+}$ : Electronic configuration is [Ar] $3d^5$
$Mn^{2+}$ : Electronic configuration is [Ar] $3d^5$

Step 3: Determine the presence of unpaired electrons.
$Mn^{7+}$ ( $d^0$ ): No unpaired electrons.
$Mn^{6+}$ ( $d^1$ ): One unpaired electron.
$Mn^{2+}$ ( $d^5$ ): Five unpaired electrons.
$Mn^{2+}$ ( $d^5$ ): Five unpaired electrons.

Step 4: Determine if the compound is colored.
$\mathrm{KMnO}_4$ is purple in color. Although it has no unpaired d-electrons, it exhibits color due to charge transfer transitions.
$\mathrm{K}_2\mathrm{MnO}_4$ is green in color and has one unpaired electron. The color is due to d-d transitions.
$\mathrm{MnSO}_4$ is pale pink in color and has five unpaired electrons. The color is due to d-d transitions.
$\mathrm{MnCl}_2$ is also pale pink and has five unpaired electrons. The color is due to d-d transitions.

Step 5: Identify the compound with color but no unpaired electrons.
$\mathrm{KMnO}_4$ has color but no unpaired electrons.

Final Answer: $\mathrm{KMnO}_4$

Why other options are wrong:
- **Option B ( $\mathrm{K}_2 \mathrm{MnO}_4$ ): This is incorrect because it has one unpaired electron.
- Option C ( $\mathrm{MnSO}_4$ ): This is incorrect because it has five unpaired electrons.
- Option D ( $\mathrm{MnCl}_2$ ): This is incorrect because it has five unpaired electrons.

Key Concepts:
- Charge transfer transitions
- d-d transitions
- Electronic configuration
- Oxidation states
- Color of coordination compounds

Memory Aids:
- Charge transfer transitions can occur even in the absence of d electrons.

Quick Tips:**
- Remember that color in coordination compounds can arise from both d-d transitions and charge transfer transitions.
- Compounds with $d^0$ or $d^{10}$ configurations can still be colored due to charge transfer transitions.

71 Easy

71 of 80Question permalink

Remembering +4 | -1

The magnetic moment

\mathrm{\mu }

, of transition metals is related to the number of unpaired elelctrnos

n

as

View answer and explanation

Correct option: D

\mathrm{\mu } = \sqrt {n\left( {n + 2} \right)}

Option D is correct because the magnetic moment is calculated using the spin-only formula involving the number of unpaired electrons.

Solution approach

Correct Answer + Core Reason: $\mathrm{\mu } = \sqrt {n {n + 2} }$ . The magnetic moment is calculated using the spin-only formula.

Step-by-Step Solution:

Step 1: Identify the formula for calculating the magnetic moment ( $\mu$ ) of transition metals based on the number of unpaired electrons (n).

Step 2: The formula is given by: $\mu = \sqrt{n(n+2)}$ Bohr magnetons.

Step 3: This formula arises from considering only the spin angular momentum contribution to the magnetic moment. The orbital angular momentum contribution is often quenched in transition metal complexes.

Why other options are wrong:
- Option A: $\mathrm{\mu } = n{\left( {n + 2} \right)^2}$ is incorrect because it squares the (n+2) term instead of taking the square root of n(n+2).
- Option B: $\mathrm{\mu } = {n^2}\left( {n + 2} \right)$ is incorrect because it squares 'n' and does not take the square root of the entire expression.
- Option C: $\mathrm{\mu } = \frac{n}{{\left( {n + 2} \right)}}$ is incorrect because it divides 'n' by (n+2) instead of using the correct square root formula.

Key Concepts:
- Magnetic moment of transition metals
- Spin-only formula
- Unpaired electrons

Memory Aids:
- Remember the formula: $\mu = \sqrt{n(n+2)}$

Quick Tips:
- The number of unpaired electrons determines the magnetic properties of transition metal ions. Use the correct formula to calculate the magnetic moment.

72 Easy

72 of 80Question permalink

Understanding +4 | -1

Which will reduce acidified potassium dichromate solution?

View answer and explanation

Correct option: B

Mohr’s salt

Option B (Mohr’s salt) is correct because Mohr's salt (ferrous ammonium sulfate) contains

Fe^{2+}

ions, which can be oxidized to

Fe^{3+}

by potassium dichromate, thus reducing the dichromate.

Solution approach

Correct Answer + Core Reason: Mohr’s salt. Mohr's salt contains $Fe^{2+}$ ions, which can reduce dichromate.

Step-by-Step Solution:

Step 1: Recall the composition of the given salts.

Step 2: Potash alum: ${\mathrm{KAl}{{\left( {\mathrm{SO}_4} \right)}_2}.12{\mathrm{H}_2}\mathrm{O}}$ \nMohr’s salt: ${{\left( {\mathrm{NH}_4} \right)}_2}\mathrm{Fe}{{\left( {\mathrm{SO}_4} \right)}_2}.6{\mathrm{H}_2}\mathrm{O}$ \nChile saltpetre: $\mathrm{NaN}{\mathrm{O}_3}$ \nWhite vitriol: $\mathrm{ZnSO}_4}.7{\mathrm{H}_2}\mathrm{O}$

Step 3: Potassium dichromate is an oxidizing agent. It can be reduced by a reducing agent.

Step 4: Mohr’s salt contains $Fe^{2+}$ ions, which can be oxidized to $Fe^{3+}$ ions, thus acting as a reducing agent.

Why other options are wrong:
- Option A: Potash alum: Aluminum is already in its highest oxidation state (+3) and cannot be further oxidized.
- Option C: Chile saltpetre: Sodium nitrate is an oxidizing agent itself and cannot reduce potassium dichromate.
- Option D: White vitriol: Zinc is not easily oxidized and does not act as a reducing agent in this case.

Key Concepts:
- Redox reactions
- Oxidizing and reducing agents
- Potassium dichromate
- Mohr’s salt

Memory Aids:
- Mohr's salt contains $Fe^{2+}$ , which is easily oxidized.

Quick Tips:
- Identify the oxidation states of the elements in the compounds to determine their reducing or oxidizing capabilities.

73 Easy

73 of 80Question permalink

Remembering +4 | -1

Brass is an alloy of:

View answer and explanation

Correct option: A

Zn and Cu

Option A (Zn and Cu) is correct because brass is an alloy primarily composed of zinc and copper.

Solution approach

Correct Answer + Core Reason: Zn and Cu. Brass is an alloy of zinc and copper.

Step-by-Step Solution:

Step 1: Recall the composition of common alloys.

Step 2: Brass is an alloy of copper (Cu) and zinc (Zn).

Why other options are wrong:
- Option B (Cu and Sn): Copper and tin form bronze.
- Option C (Zn and Sn): Zinc and tin are not the primary components of brass.
- Option D (Cu, Zn and sn): Tin is not a primary component of brass.

Key Concepts:
- Composition of alloys
- Brass composition

Memory Aids:
- Brass = Copper + Zinc.

Quick Tips:
- Remember the compositions of common alloys.

74 Easy

74 of 80Question permalink

Application +4 | -1

Which of the following in its oxidation state shows the paramagnetism?

View answer and explanation

Correct option: A

\mathrm{Tb}(\mathrm{IV})

Option A (

\mathrm{Tb}(\mathrm{IV})

) is correct because

\mathrm{Tb}(\mathrm{IV})

has unpaired electrons, making it paramagnetic.

Solution approach

Correct Answer + Core Reason: $\mathrm{Tb}(\mathrm{IV})$ . $\mathrm{Tb}(\mathrm{IV})$ has unpaired electrons.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of each ion.

Step 2: $\mathrm{Tb}$ (Z=65): $[\mathrm{Xe}] 4f^9 6s^2$ . $\mathrm{Tb}^{4+}$ : $[\mathrm{Xe}] 4f^7$ . 7 unpaired electrons, paramagnetic.

Step 3: $\mathrm{Lu}$ (Z=71): $[\mathrm{Xe}] 4f^{14} 6s^2$ . $\mathrm{Lu}^{3+}$ : $[\mathrm{Xe}] 4f^{14}$ . All electrons paired, diamagnetic.

Step 4: $\mathrm{Ce}$ (Z=58): $[\mathrm{Xe}] 4f^1 5d^1 6s^2$ . $\mathrm{Ce}^{4+}$ : $[\mathrm{Xe}]$ . All electrons paired, diamagnetic.

Step 5: $\mathrm{La}$ (Z=57): $[\mathrm{Xe}] 5d^1 6s^2$ . $\mathrm{La}^{3+}$ : $[\mathrm{Xe}]$ . All electrons paired, diamagnetic.

Step 6: Only $\mathrm{Tb}^{4+}$ has unpaired electrons and is paramagnetic.

Why other options are wrong:
- Option B: $\mathrm{Lu}(\mathrm{III})$ has a completely filled f-orbital and is diamagnetic.
- Option C: $\mathrm{Ce}(\mathrm{IV})$ has no f-electrons and is diamagnetic.
- Option D: $\mathrm{La}(\mathrm{III})$ has no f-electrons and is diamagnetic.

Key Concepts:
- Electronic configuration
- Paramagnetism and diamagnetism
- Hund's rule

Memory Aids:
- Paramagnetic substances have unpaired electrons.

Quick Tips:
- Determine the number of unpaired electrons to predict magnetic properties.

75 Easy

75 of 80Question permalink

Analyzing +4 | -1

The correct order of spin only magnetic moments for the following complex ions is

View answer and explanation

Correct option: C

\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-} < \left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{3-}< \left[\mathrm{CoF}_{6}\right]^{3-} < \left[\mathrm{MnBr}_{4}\right]^{2-}

Option C (

[Fe(CN)_6]^{3-} < [Mn(CN)_6]^{3-} < [CoF_6]^{3-} < [MnBr_4]^{2-}

) is correct because it arranges the complexes in increasing order of their spin-only magnetic moments, which depends on the number of unpaired electrons.

Solution approach

Correct Answer + Core Reason: $[\mathrm{Fe}(\mathrm{CN})_{6}]^{3-} < [\mathrm{Mn}(\mathrm{CN})_{6}]^{3-}< [\mathrm{CoF}_{6}]^{3-} < [\mathrm{MnBr}_{4}]^{2-}$ . The order is determined by the number of unpaired electrons in each complex.

Step-by-Step Solution:

Step 1: Determine the oxidation state of the metal ion in each complex.
$[Fe(CN)_6]^{3-}$ : Fe is in +3 oxidation state.
$[Mn(CN)_6]^{3-}$ : Mn is in +3 oxidation state.
$[CoF_6]^{3-}$ : Co is in +3 oxidation state.
$[MnBr_4]^{2-}$ : Mn is in +2 oxidation state.

Step 2: Determine the electronic configuration of the metal ion in each complex.
$Fe^{3+}$ : Electronic configuration is $d^5$ . $CN^-$ is a strong field ligand, so it will pair up the electrons. Thus, the configuration becomes $t_{2g}^5 e_g^0$ , with 1 unpaired electron.
$Mn^{3+}$ : Electronic configuration is $d^4$ . $CN^-$ is a strong field ligand, so it will pair up the electrons. Thus, the configuration becomes $t_{2g}^4 e_g^0$ , with 2 unpaired electrons.
$Co^{3+}$ : Electronic configuration is $d^6$ . $F^-$ is a weak field ligand, so it will not pair up the electrons. Thus, the configuration becomes $t_{2g}^4 e_g^2$ , with 4 unpaired electrons.
$Mn^{2+}$ : Electronic configuration is $d^5$ . $Br^-$ is a weak field ligand, so it will not pair up the electrons. Thus, the configuration becomes $t_{2g}^3 e_g^2$ , with 5 unpaired electrons.

Step 3: Calculate the spin-only magnetic moment using the formula $\mu = \sqrt{n(n+2)}$ BM, where n is the number of unpaired electrons.
$[Fe(CN)_6]^{3-}$ : n = 1, $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73$ BM
$[Mn(CN)_6]^{3-}$ : n = 2, $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83$ BM
$[CoF_6]^{3-}$ : n = 4, $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90$ BM
$[MnBr_4]^{2-}$ : n = 5, $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92$ BM

Step 4: Arrange the complexes in increasing order of their spin-only magnetic moments.
$[Fe(CN)_6]^{3-} < [Mn(CN)_6]^{3-} < [CoF_6]^{3-} < [MnBr_4]^{2-}$

Final Answer: $[\mathrm{Fe}(\mathrm{CN})_{6}]^{3-} < [\mathrm{Mn}(\mathrm{CN})_{6}]^{3-}< [\mathrm{CoF}_{6}]^{3-} < [\mathrm{MnBr}_{4}]^{2-}$

Why other options are wrong:
- Option A: This is incorrect because it does not correctly account for the effect of strong and weak field ligands on the number of unpaired electrons.
- Option B: This is incorrect because it does not correctly account for the effect of strong and weak field ligands on the number of unpaired electrons.
- Option D: This is incorrect because it does not correctly account for the effect of strong and weak field ligands on the number of unpaired electrons.

Key Concepts:
- Spin-only magnetic moment
- Crystal field theory
- Strong and weak field ligands
- Electronic configuration
- Oxidation states

Memory Aids:
- Strong field ligands cause pairing of electrons, leading to lower spin complexes.
- Weak field ligands do not cause pairing of electrons, leading to high spin complexes.

Quick Tips:
- Determine the oxidation state and electronic configuration of the metal ion first.
- Identify the ligands as strong or weak field ligands.
- Determine the number of unpaired electrons based on the ligand field strength.
- Calculate the spin-only magnetic moment using the formula $\mu = \sqrt{n(n+2)}$ BM.

76 Easy

76 of 80Question permalink

Remembering +4 | -1

The nitrate of which metal leaves metallic globule on heating strongly?

View answer and explanation

Correct option: B

\mathrm{AgN}{\mathrm{O}_3}

Option B (

\mathrm{AgN}{\mathrm{O}_3}

) is correct because silver nitrate decomposes upon heating to produce metallic silver, nitrogen dioxide, and oxygen.

Solution approach

Correct Answer + Core Reason: $\mathrm{AgN}{\mathrm{O}_3}$ . Silver nitrate decomposes to form metallic silver on heating.

Step-by-Step Solution:

Step 1: Consider the decomposition of each nitrate upon heating.

Step 2: $\mathrm{Cu}{\left( {\mathrm{N}{\mathrm{O}_3}} \right)_2}$ decomposes to form copper oxide (CuO), not metallic copper.

Step 3: $\mathrm{AgN}{\mathrm{O}_3}$ decomposes to form metallic silver (Ag), nitrogen dioxide ( $NO_2$ ), and oxygen ( $O_2$ ). The reaction is: $2AgNO_3(s) \rightarrow 2Ag(s) + 2NO_2(g) + O_2(g)$

Step 4: $\mathrm{NaN}{\mathrm{O}_3}$ decomposes to form sodium nitrite ( $NaNO_2$ ) and oxygen.

Step 5: $\mathrm{Pb}{\left( {\mathrm{N}{\mathrm{O}_3}} \right)_2}$ decomposes to form lead oxide (PbO), not metallic lead.

Step 6: Therefore, $\mathrm{AgN}{\mathrm{O}_3}$ leaves a metallic globule (Ag) on heating.

Why other options are wrong:
- Option A: Copper nitrate decomposes to copper oxide, not metallic copper.
- Option C: Sodium nitrate decomposes to sodium nitrite, not metallic sodium.
- Option D: Lead nitrate decomposes to lead oxide, not metallic lead.

Key Concepts:
- Decomposition Reactions
- Metal Nitrates

Memory Aids:
- Silver nitrate gives silver metal on heating.

Quick Tips:
- Remember the decomposition products of common metal nitrates.

77 Easy

77 of 80Question permalink

Analyzing +4 | -1

Which of the following is most stable?

View answer and explanation

Correct option: D

\mathrm{C}{\mathrm{r}^{3 + }}

Option D (

\mathrm{C}{\mathrm{r}^{3 + }}

) is the most stable because it has a half-filled

t_{2g}

orbital configuration.

Solution approach

Correct Answer + Core Reason: $\mathrm{C}{\mathrm{r}^{3 + }}$ is the most stable due to its half-filled $t_{2g}$ configuration.

Step-by-Step Solution:

Step 1: Determine the electronic configurations of the given ions.
- ${\mathrm{V}^{3 + }}$ : $[\mathrm{Ar}]3{d^2}$
- $\mathrm{T}{\mathrm{i}^{3 + }}$ : $[\mathrm{Ar}]3{d^1}$
- $\mathrm{M}{\mathrm{n}^{3 + }}$ : $[\mathrm{Ar}]3{d^4}$
- $\mathrm{C}{\mathrm{r}^{3 + }}$ : $[\mathrm{Ar}]3{d^3}$

Step 2: Consider the effect of crystal field splitting in octahedral complexes. The $d$ orbitals split into $t_{2g}$ and $e_g$ sets.

Step 3: Evaluate the stability based on the filling of $t_{2g}$ orbitals. Half-filled ( $t_{2g}^3$ ) and fully filled ( $t_{2g}^6$ ) configurations are particularly stable.

Step 4: $\mathrm{C}{\mathrm{r}^{3 + }}$ has a $d^3$ configuration, which corresponds to a half-filled $t_{2g}$ set ( $t_{2g}^3$ ). This makes it more stable than the other options.

Why other options are wrong:
- **Option A ( $\mathrm{V}^{3 + }$ ):** The $d^2$ configuration is not particularly stable.
- **Option B ( $\mathrm{T}{\mathrm{i}^{3 + }}$ ):** The $d^1$ configuration is not particularly stable.
- **Option C ( $\mathrm{M}{\mathrm{n}^{3 + }}$ ):** The $d^4$ configuration is not as stable as a half-filled $t_{2g}$ configuration. It also has a tendency to undergo Jahn-Teller distortion.

Key Concepts:
- Electronic configuration of transition metal ions
- Crystal field theory
- Stability of half-filled and fully filled orbitals

Memory Aids:
- Remember that half-filled and fully filled orbitals generally lead to increased stability.

Quick Tips:
- When comparing the stability of transition metal ions, consider their electronic configurations and the effect of crystal field splitting.

78 Easy

78 of 80Question permalink

Understanding +4 | -1

4{\mathrm{K}_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7}\xrightarrow{{\mathrm{Heat}}}4{\mathrm{K}_2}\mathrm{Cr}{\mathrm{O}_4} + 3{\mathrm{O}_2} + X

In the above reaction,

X

is

View answer and explanation

Correct option: C

\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_3}

Option C (

\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_3}

) is correct because when potassium dichromate is heated, it decomposes to form potassium chromate, oxygen, and chromium(III) oxide.

Solution approach

Correct Answer + Core Reason: $\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_3}$ . Heating potassium dichromate produces chromium(III) oxide.

Step-by-Step Solution:

Step 1: Recall the thermal decomposition of potassium dichromate.

Step 2: The balanced equation for the decomposition is:
$4{\mathrm{K}_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7}\xrightarrow{{\mathrm{Heat}}}4{\mathrm{K}_2}\mathrm{Cr}{\mathrm{O}_4} + 2\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_3} + 3{\mathrm{O}_2}$
(Note: Original equation had a typo. It should be 2Cr2O3, not 3O2 + X)

Step 3: Identify X as $\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_3}$ .

Why other options are wrong:
- Option A: $\mathrm{Cr}{\mathrm{O}_3}\mathrm{\;}$ : Chromium trioxide is not formed in this decomposition.
- Option B: $\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7}$ : Dichromate is the reactant, not a product.
- Option D: $\mathrm{Cr}{\mathrm{O}_5}$ : Chromium pentoxide is not formed in this decomposition.

Key Concepts:
- Thermal decomposition
- Potassium dichromate

Memory Aids:
- Dichromate decomposes to chromate, chromium(III) oxide, and oxygen.

Quick Tips:
- Remember the products of common decomposition reactions.

79 Easy

79 of 80Question permalink

Analyzing +4 | -1

The set of correct statements is:

(i) Manganese exhibits +7 oxidation state in its oxide.

(ii) Ruthenium and Osmium exhibit +8 oxidation in their oxides.

(iii) Sc shows +4 oxidation state which is oxidizing in nature.

(iv) Cr shows oxidising nature in +6 oxidation state.

View answer and explanation

Correct option: B

(i), (ii) and (iv)

Option B ((i), (ii) and (iv)) is correct because Manganese exhibits +7 in its oxide, Ruthenium and Osmium exhibit +8 in their oxides, and Cr shows oxidizing nature in +6 oxidation state.

Solution approach

Correct Answer + Core Reason: (i), (ii) and (iv). These statements are correct about the oxidation states and properties of the given elements.

Step-by-Step Solution:

Step 1: Analyze statement (i):\nManganese exhibits +7 oxidation state in its oxide. This is true, as seen in $Mn_2O_7$ .

Step 2: Analyze statement (ii):\nRuthenium and Osmium exhibit +8 oxidation in their oxides. This is true, as seen in $RuO_4$ and $OsO_4$ .

Step 3: Analyze statement (iii):\nSc shows +4 oxidation state which is oxidizing in nature. Scandium primarily exhibits a +3 oxidation state. While Sc can exhibit +4 oxidation state, it is not common and not oxidizing in nature. This statement is incorrect.

Step 4: Analyze statement (iv):\nCr shows oxidizing nature in +6 oxidation state. This is true, as seen in $Cr_2O_7^{2-}$ and $CrO_4^{2-}$ , which are strong oxidizing agents.

Step 5: The correct statements are (i), (ii), and (iv).

Why other options are wrong:
- Option A: Includes statement (iii), which is incorrect.
- Option C: Includes statement (iii), which is incorrect.
- Option D: Does not include statement (ii), which is correct.

Key Concepts:
- Oxidation states of transition metals
- Oxidizing agents

Memory Aids:
- Transition metals exhibit variable oxidation states.

Quick Tips:
- Know the common oxidation states of transition metals.

80 Easy

80 of 80Question permalink

Application +4 | -1

In chromite ore, the oxidation number of iron and chromium are respectively.

View answer and explanation

Correct option: D

+ 2, + 3

Option D is correct because in chromite ore (FeCr2O4), iron has a +2 oxidation state and chromium has a +3 oxidation state.

Solution approach

Correct Answer + Core Reason: $+ 2, + 3$ . In chromite ore (FeCr₂O₄), iron is +2 and chromium is +3.

Step-by-Step Solution:

Step 1: Write the formula for chromite ore.\nThe formula for chromite ore is FeCr₂O₄.

Step 2: Determine the oxidation state of oxygen.\nThe oxidation state of oxygen is -2.

Step 3: Let the oxidation state of iron be x and the oxidation state of chromium be y.\nThen, the equation for the oxidation states is:\nx + 2y + 4(-2) = 0\nx + 2y - 8 = 0\nx + 2y = 8

Step 4: Consider the possible oxidation states of iron and chromium.\nIron commonly exhibits +2 and +3 oxidation states. Chromium commonly exhibits +3 and +6 oxidation states.

Step 5: Test the possible combinations.\nIf x = +2, then 2 + 2y = 8 => 2y = 6 => y = +3. This is a possible combination.\nIf x = +3, then 3 + 2y = 8 => 2y = 5 => y = +2.5. This is not a likely oxidation state for chromium.

Step 6: Conclude the answer.\nThe oxidation state of iron is +2, and the oxidation state of chromium is +3. Therefore, the correct answer is +2, +3.

Why other options are wrong:
- Option A: +3, +2 is incorrect because iron is +2 and chromium is +3 in chromite ore.
- Option B: +3, +6 is incorrect because chromium is not +6 in chromite ore.
- Option C: +2, +6 is incorrect because chromium is not +6 in chromite ore.

Key Concepts:
- Oxidation state
- Chromite ore
- Transition metal compounds

Memory Aids:
- Remember the formula for chromite ore: FeCr₂O₄.

Quick Tips:
- Use the rules for assigning oxidation states to determine the oxidation state of each element in the compound.

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