${\rm{NaN}}{{\rm{H}}_{\rm{2}}}{\rm{ + }}{{\rm{N}}_{\rm{2}}}{\rm{O }} \to {\rm{X + NaOH + N}}{{\rm{H}}_{\rm{3}}}$ what is the $X$?

${\rm{N}}{{\rm{a}}_{\rm{3}}}{\rm{N}}$

Among the following statement, the correct statement about ${\rm{P}}{{\rm{H}}_3}$ and ${\rm{N}}{{\rm{H}}_3}$ is:

${\rm{N}}{{\rm{H}}_3}$ is a better electron donor because the lone pair of electron occupies spherical s-orbital and is less directional

${\rm{P}}{{\rm{H}}_3}$ is a better electron donor because the lone pair of electron occupies spherical s-orbital and is less directional

When copper is heated with conc. ${\rm{HN}}{{\rm{O}}_3}$ it produces:

${\rm{Cu(N}}{{\rm{O}}_3}{{\rm{)}}_2}\,\,{\rm{and}}\,{\rm{N}}{{\rm{O}}_2}$

${\rm{Cu(N}}{{\rm{O}}_3}{{\rm{)}}_2}\,\,{\rm{and}}\,{\rm{NO}}$

${\rm{Cu(N}}{{\rm{O}}_3}{{\rm{)}}_2},\,{\rm{NO}}\,\,{\rm{and}}\,{\rm{N}}{{\rm{O}}_2}$

${\rm{Cu(N}}{{\rm{O}}_3}{{\rm{)}}_2}\,{\rm{and}}\,{{\rm{N}}_2}{\rm{O}}$

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NEET • CHEMISTRY • P-BLOCK ELEMENTS

P-block elements practice · Medium

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Showing 61-90 of 93 questions

61 Medium

61 of 93

Application +4 | -1

C{l_2}

changed to

C{l^ - }

and

Cl{O^ - }

in cold and NaOH. Equivalent weight of

C{l_2}

will be

View answer and explanation

Correct option: A

M

The correct answer is A (

M

) because the equivalent weight of

Cl_2

in the given reaction is equal to its molecular weight due to the disproportionation reaction.

62 Medium

62 of 93

Understanding +4 | -1

The compound which has the highest reducing character among the following is

View answer and explanation

Correct option: D

{H_2}Te

Option D (

{H_2}Te

) is correct because reducing character increases down the group due to the decreasing bond dissociation energy of the H-E bond.

63 Medium

63 of 93

Application +4 | -1

\mathrm{NaN}{\mathrm{H}_\mathrm{2}}\mathrm{ + }{\mathrm{N}_\mathrm{2}}\mathrm{O } \to \mathrm{X + NaOH + N}{\mathrm{H}_\mathrm{3}}

what is the

X

View answer and explanation

Correct option: C

\mathrm{Na}{\mathrm{N}_\mathrm{3}}

Option C (

\mathrm{Na}{\mathrm{N}_\mathrm{3}}

) is correct because the reaction of sodium amide with nitrous oxide produces sodium azide, sodium hydroxide, and ammonia.

64 Medium

64 of 93

Remembering +4 | -1

A hybrid rocket propellant uses

View answer and explanation

Correct option: A

A liquid oxidizer and a solid fuel

Option A (A liquid oxidizer and a solid fuel) is correct because hybrid rockets utilize a combination of a liquid oxidizer and a solid fuel for propulsion.

65 Medium

65 of 93

Application +4 | -1

\mathrm{aCl}_2+\mathrm{bOH}^{-} \longrightarrow \mathrm{cCl}^{-}+\mathrm{dClO}^{-}+\mathrm{eH}_2 \mathrm{O}

Determine the value of

(\mathrm{a+b+c+d+e})

, if reaction is balanced with simplest integer coefficients

View answer and explanation

Correct option: A

The correct answer is A (6) because when the balanced equation has simplest integer coefficients, the sum of the coefficients a, b, c, d, and e is 6.

66 Medium

66 of 93

Analyzing +4 | -1

Which of the following oxy-acids of phosphorus is a reducing agent and monobasic?

View answer and explanation

Correct option: A

{\mathrm{H}_\mathrm{3}}\mathrm{P}{\mathrm{O}_\mathrm{2}}

Option A is correct because

{\mathrm{H}_\mathrm{3}}\mathrm{P}{\mathrm{O}_\mathrm{2}}

(Hypophosphorous acid) is a reducing agent due to the presence of two P-H bonds and is monobasic because it has one ionizable hydrogen.

67 Medium

67 of 93

Analyzing +4 | -1

When

\mathrm{C}{\mathrm{l}_2}

water is added to an aqueous solution of potassium halide in presence of chloroform a violet colour is obtained. On adding more of

\mathrm{C}{\mathrm{l}_2}

water, the violet colour disappears and a colourless solution is obtained. This test confirms the presence of the following in aqueous solution:

View answer and explanation

Correct option: A

Iodide

Option A (Iodide) is correct because the violet color in chloroform indicates the presence of iodine (

I_2

), which is formed by the oxidation of iodide ions by chlorine water.

68 Medium

68 of 93

Understanding +4 | -1

Disproportionation phenomenon is not shown by

View answer and explanation

Correct option: B

ClO_4^ -

Option B (

ClO_4^ -

) is correct because the oxidation state of Cl in

ClO_4^ -

is +7, which is the highest possible oxidation state for chlorine, so it cannot be further oxidized and thus cannot disproportionate.

69 Medium

69 of 93

Application +4 | -1

Hot conc.

{H_2}S{O_4}

act as moderately strong oxidizing agent. It oxidises both metals and non-metals. Which of the following element is oxidized by conc.

{H_2}S{O_4}

into two gaseous products.

View answer and explanation

Correct option: C

Option C (C) is correct because hot concentrated

{H_2}S{O_4}

oxidizes carbon into two gaseous products: carbon dioxide (

CO_2

) and sulfur dioxide (

SO_2

70 Medium

70 of 93

Remembering +4 | -1

Antimony dissolves in aquaregia to give:

View answer and explanation

Correct option: C

\mathrm{SbC}{\mathrm{l}_5}

Option C is correct. Antimony dissolves in aqua regia (a mixture of nitric acid and hydrochloric acid) to form antimony pentachloride (SbCl5).

71 Medium

71 of 93

Understanding +4 | -1

Which among the following does not disproportionate in alkaline medium?

View answer and explanation

Correct option: A

{F_2}

Option A (

{F_2}

) is correct because fluorine is the strongest oxidizing agent and only undergoes reduction, not disproportionation in alkaline medium.

72 Medium

72 of 93

Remembering +4 | -1

Complex formed in brown ring test for nitrates is

View answer and explanation

Correct option: A

{\left[ {Fe{{\left( {{H_2}O} \right)}_5}\left( {NO} \right)} \right]^{2 + }}

Option A (

{\left[ {Fe{{\left( {{H_2}O} \right)}_5}\left( {NO} \right)} \right]^{2 + }}

) is correct because the brown ring complex is formed when ferrous ions react with nitrate ions in the presence of sulfuric acid, resulting in the pentaaquanitrosyliron(II) ion.

73 Medium

73 of 93

Remembering +4 | -1

Which of the following represents a biliquid propellant?

View answer and explanation

Correct option: C

{\mathrm{N}_2}{\mathrm{O}_4}\, +

Unsymmetrical dimethyl hydrazene

Option C (

{\mathrm{N}_2}{\mathrm{O}_4}\, +

Unsymmetrical dimethyl hydrazene) is correct because it represents a common combination of oxidizer and fuel used in biliquid propellant systems.

74 Medium

74 of 93

Understanding +4 | -1

Among the following statement, the correct statement about

\mathrm{P}{\mathrm{H}_3}

and

\mathrm{N}{\mathrm{H}_3}

is:

View answer and explanation

Correct option: C

\mathrm{N}{\mathrm{H}_3}

is a better electron donor because the lone pair of electron occupies

s{p^3}

orbital and more directional

Option C (

\mathrm{N}{\mathrm{H}_3}

is a better electron donor because the lone pair of electron occupies

s{p^3}

orbital and more directional) is correct because nitrogen is more electronegative than phosphorus, leading to a more directional and available lone pair in

N{H_3}

75 Medium

75 of 93

Analyzing +4 | -1

The strongest acid is:

View answer and explanation

Correct option: C

{\mathrm{H}_4}{\mathrm{P}_2}{\mathrm{O}_7}

Option C is correct because

{\mathrm{H}_4}\mathrm{P}_2{\mathrm{O}_7}

(Pyrophosphoric acid) is the strongest acid among the given options due to its structure and the number of ionizable protons.

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76 Medium

76 of 93

Understanding +4 | -1

The correct order of reducing power of halide ions is :

View answer and explanation

Correct option: D

{\mathrm{I}^ - } > B{\mathrm{r}^ - } > \mathrm{C}{\mathrm{l}^ - } > {\mathrm{F}^ - }

Option D (

{\mathrm{I}^ - } > B{\mathrm{r}^ - } > \mathrm{C}{\mathrm{l}^ - } > {\mathrm{F}^ - }

) is correct because reducing power increases down the group due to increasing size and decreasing electron gain enthalpy.

77 Medium

77 of 93

Analyzing +4 | -1

Chemical nature of the nitrogen oxide compound obtained from a reaction of concentrated nitric acid and

{\mathrm{P}_4}{\mathrm{O}_{10}}

(in 4: 1 ratio) is

View answer and explanation

Correct option: A

acidic

Option A (acidic) is correct because the reaction between concentrated nitric acid and

P_4O_{10}

produces

N_2O_5

, which is the anhydride of nitric acid and thus acidic in nature.

78 Medium

78 of 93

Understanding +4 | -1

A green yellow gas reacts with an alkali metal hydroxide to form a halate which can used in fireworks and safety matches. The gas and halate respectively are

View answer and explanation

Correct option: B

C{l_2},\;KCl{O_3}

Option B (

Cl_2, KClO_3

) is correct because chlorine gas reacts with alkali metal hydroxide to form a chlorate, such as potassium chlorate, which is used in fireworks and safety matches.

79 Medium

79 of 93

Remembering +4 | -1

Nitric acid is generally light yellow due to the presence of:

View answer and explanation

Correct option: C

\mathrm{N}{\mathrm{O}_2}

Option C (NO2) is correct because nitric acid decomposes slightly, especially upon exposure to light, forming nitrogen dioxide, which imparts a yellow color.

80 Medium

80 of 93

Understanding +4 | -1

When copper is heated with conc.

\mathrm{HN}{\mathrm{O}_3}

it produces:

View answer and explanation

Correct option: A

\mathrm{Cu(N}{\mathrm{O}_3}{\mathrm{)}_2}\,\,\mathrm{and}\,\mathrm{N}{\mathrm{O}_2}

Option A (

\mathrm{Cu(N}{\mathrm{O}_3}{\mathrm{)}_2}\,\mathrm{and}\,\mathrm{N}{\mathrm{O}_2}

) is correct because the reaction of copper with concentrated nitric acid produces copper nitrate and nitrogen dioxide.

81 Medium

81 of 93

Analyzing +4 | -1

Which of the following statements is wrong?

View answer and explanation

Correct option: A

Single N – N bond is stronger than the single P – P bond

Option A is correct because the single N-N bond is weaker than the single P-P bond due to smaller size and higher interelectronic repulsion in nitrogen.

82 Medium

82 of 93

Remembering +4 | -1

Fluorine gas is stored in:

View answer and explanation

Correct option: C

Both (1) and (2)

Option C (Both (1) and (2)) is correct because fluorine gas can be stored in both copper and steel vessels due to the formation of a protective fluoride layer on the metal surface.

83 Medium

83 of 93

Understanding +4 | -1

The correct shape and hybridization for

Xe{F_2}

are

View answer and explanation

Correct option: C

Linear,s{p^3}d

Option C (

Linear, sp^3d

) is correct because

XeF_2

has a linear shape with three lone pairs on Xe, resulting in

sp^3d

hybridization.

84 Medium

84 of 93

Application +4 | -1

Which of the following is linear?

View answer and explanation

Correct option: A

\mathrm{XeF}_2

Option A (

\mathrm{XeF}_2

) is correct because it has a linear structure due to the arrangement of bonding pairs and lone pairs around the central Xe atom.

85 Medium

85 of 93

Remembering +4 | -1

Dinitrogen pentoxide a colourless solid is prepared by

View answer and explanation

Correct option: C

Dehydrating

\mathrm{HN}{\mathrm{O}_3}

with

{\mathrm{P}_\mathrm{4}}{\mathrm{O}_{\mathrm{10}}}

Option C is correct because dinitrogen pentoxide is prepared by dehydrating nitric acid with phosphorus pentoxide.

86 Medium

86 of 93

Remembering +4 | -1

Orthophosphoric acid is ionized in…….steps.

View answer and explanation

Correct option: C

Option C (3) is correct because orthophosphoric acid (H3PO4) is a triprotic acid, meaning it can donate three protons (H+) in three ionization steps.

87 Medium

87 of 93

Application +4 | -1

Find the number of lone pairs of electrons on

\mathrm{Xe}

\mathrm{XeOF}_4

View answer and explanation

Correct option: A

Option A (1) is correct because in

\mathrm{XeOF}_4

, Xenon has one lone pair of electrons.

88 Medium

88 of 93

Analyzing +4 | -1

Which statement is wrong?

View answer and explanation

Correct option: C

Both show +2, +4 and +6 oxidation states

Option C (Both show +2, +4 and +6 oxidation states) is wrong because Oxygen does not show +2, +4 and +6 oxidation states.

89 Medium

89 of 93

Application +4 | -1

In the reaction,

\mathrm{MnO}_4^ - \mathrm{\;} + \mathrm{\;}{\mathrm{I}^ - }\;\mathop \to \limits^{A\mathrm{lkaline\;solution}} \;\left[ X \right];\,\,\left[ X \right]\;\mathrm{is}:

View answer and explanation

Correct option: A

\mathrm{IO}_3^ -

Option A (

\mathrm{IO}_3^ -

) is correct because in alkaline solution, permanganate ions oxidize iodide ions to iodate ions.

90 Medium

90 of 93

Application +4 | -1

Xe{O_3},\;Xe\;

is:

View answer and explanation

Correct option: A

s{p^3} -

hybridized

Option A (

sp^3

- hybridized) is correct because Xe in

XeO_3

has three sigma bonds and one lone pair, leading to

sp^3

hybridization.

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