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

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NEET • CHEMISTRY • P-BLOCK ELEMENTS

P-block elements practice · Mixed difficulty

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181 Hard

181 of 248

Analyzing +4 | -1

Assertion:

\mathrm{SO}_2

(at. room temp.) and

\mathrm{SeO}_2

(at higher temperature vapour condition) have same shape.

Reason: Covalent radius of

\mathrm{Se}

is some what oxidising agent. larger than sulphur.

View answer and explanation

Correct option: B

If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion

Option B is correct. SO2 has a bent shape at room temperature, and SeO2 also has a bent shape in the vapor phase at higher temperatures. The covalent radius difference is true but doesn't directly explain the shape similarity.

182 Hard

182 of 248

Analyzing +4 | -1

Assertion (A): The first real compound of the noble gases in 1962 was

{O}_2^{+}\left[{PtF}_6\right]^{-}

Reason (R): The discovery was based on the basis of comparable ionisation energy of

{O}_2

and

{Xe}

View answer and explanation

Correct option: A

Both (A) and (R) are true and (R) is the correct explanation of (A)

Option A (Both (A) and (R) are true and (R) is the correct explanation of (A)) is correct because Bartlett's discovery of

O_2^+[PtF_6]^-

led him to realize that Xenon, with a similar ionization energy, could also react with

PtF_6

183 Hard

183 of 248

Understanding +4 | -1

Which does not give ammonia with water?

View answer and explanation

Correct option: D

Ca{\left( {CN} \right)_2}

Option D (

Ca{\left( {CN} \right)_2}

) is correct because calcium cyanide (

Ca(CN)_2

) does not react with water to produce ammonia (

NH_3

184 Easy

184 of 248

Application +4 | -1

The oxidation state of N is highest in:

View answer and explanation

Correct option: A

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

Option A (

N_3H

) is correct because the oxidation state of N in hydrazoic acid (

N_3H

) is -1/3, which is higher than in the other options.

185 Easy

185 of 248

Application +4 | -1

The dihydrogen phosphate of certain metal has formula

M{H_2}P{O_4}.

The formula of metal chloride would be:

View answer and explanation

Correct option: A

MCl

Option A (

MCl

) is correct because the dihydrogen phosphate

M{H_2}P{O_4}

indicates that the metal M has a +1 charge, thus forming

MCl

186 Medium

186 of 248

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.

187 Hard

187 of 248

Analyzing +4 | -1

Assertion: Fluorine oxidises

\mathrm{H}_2 \mathrm{O}

\mathrm{O}_2

but

\mathrm{Cl}_2

does not.

Reason: Electronegativity of fluorine is greater than oxygen but EN of chlorine is lesser than oxygen.

View answer and explanation

Correct option: B

If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion

Option B is correct. Fluorine does oxidize H2O to O2, and chlorine does not under similar conditions. However, the reason given about electronegativity is not the direct explanation for this difference in oxidizing power.

188 Medium

188 of 248

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.

189 Medium

189 of 248

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.

190 Hard

190 of 248

Analyzing +4 | -1

Assertion: Salts of

\mathrm{H}_3 \mathrm{PO}_3

and

\mathrm{H}_3 \mathrm{PO}_4

used as a fertilizers.

Reason: Formed

\mathrm{H}_3 \mathrm{PO}_3

increases solubility of water in soil on addition of phosphate fertilizers.

View answer and explanation

Correct option: A

If both Assertion & Reason are True & the Reason is correct explanation of the Assertion

Option B (If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion) is correct because both the assertion and reason are true, but the reason does not explain the assertion.

191 Medium

191 of 248

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.

192 Medium

192 of 248

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.

193 Hard

193 of 248

Understanding +4 | -1

CAN pellets are coated with calcium silicate because:

View answer and explanation

Correct option: B

CAN is hygroscopic

Option B (CAN is hygroscopic) is correct because Calcium Ammonium Nitrate (CAN) absorbs moisture from the air, and the calcium silicate coating prevents this.

194 Easy

194 of 248

Analyzing +4 | -1

Which has the same electronic configuration as of noble gas?

View answer and explanation

Correct option: D

T{i^{4 + }}

Option D (

T{i^{4 + }}

) is correct because after losing 4 electrons, Titanium achieves the electronic configuration of Argon, a noble gas.

195 Hard

195 of 248

Application +4 | -1

Which of the following salt would give

S{O_2}

with hot and dil.

{H_2}S{O_4}

and also decolourises

B{r_2}

water?

View answer and explanation

Correct option: A

N{a_2}S{O_3}

Option A (

N{a_2}S{O_3}

) is correct because it reacts with hot dilute

{H_2}S{O_4}

to produce

S{O_2}

and also decolourises

B{r_2}

water due to its reducing nature.

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196 Medium

196 of 248

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.

197 Medium

197 of 248

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

198 Easy

198 of 248

Application +4 | -1

Number of lone pairs of electrons on

\mathrm{Xe}

atoms in

\mathrm{Xe}{\mathrm{F}_2},\mathrm{\;Xe}{\mathrm{F}_4}\mathrm{\;and\;Xe}{\mathrm{O}_3}

molecule are respectively

View answer and explanation

Correct option: A

3, 2 and 1

Option A (3, 2 and 1) is correct because

XeF_2

has 3 lone pairs,

XeF_4

has 2 lone pairs, and

XeO_3

has 1 lone pair on the Xe atom.

199 Medium

199 of 248

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).

200 Easy

200 of 248

Understanding +4 | -1

Decreasing acid strengths of

\mathrm{HI},\mathrm{HBr},\mathrm{HCl}

  and

HF

is:

View answer and explanation

Correct option: B

\mathrm{HI} > HBr > HCl > HF

Option B (

\mathrm{HI} > HBr > HCl > HF

) is correct because acid strength increases down the group due to decreasing bond dissociation enthalpy.

201 Easy

201 of 248

Understanding +4 | -1

The acidic character order for given oxy-acids of halogens is:

View answer and explanation

Correct option: A

\mathrm{HOCl} > HOBr > HO

Option A (

\mathrm{HOCl} > HOBr > HOI

) is correct because the acidic character of hypohalous acids decreases down the group due to decreasing electronegativity of the halogen.

202 Easy

202 of 248

Understanding +4 | -1

Which of the following is the correct order of increasing oxidising power of halogens

View answer and explanation

Correct option: A

{I_2} < B{r_2} < C{l_2} < {F_2}

Option A (

{I_2} < B{r_2} < C{l_2} < {F_2}

) is correct because oxidizing power increases up the halogen group due to increasing electronegativity.

203 Medium

203 of 248

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.

204 Easy

204 of 248

Remembering +4 | -1

The substance used as a smokescreen in war fare is

View answer and explanation

Correct option: B

P{H_3}

Option A (

SiC{l_4}

) is correct because silicon tetrachloride reacts with moisture in the air to produce HCl fumes and silica, creating a dense white smoke.

205 Easy

205 of 248

Analyzing +4 | -1

Which of the following properties does not correspond to the order?

HI < HBr < HCl < HF

View answer and explanation

Correct option: B

Reducing power

Option B (Reducing power) is correct because the reducing power increases down the group, so the order should be

HF < HCl < HBr < HI

206 Medium

206 of 248

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.

207 Medium

207 of 248

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.

208 Hard

208 of 248

Analyzing +4 | -1

Which of the following dissolves in water but does not give any oxyacid solution?

View answer and explanation

Correct option: B

O{F_2}

Option B (

O{F_2}

) is correct because

OF_2

reacts with water to produce

O_2

and

HF

, not an oxyacid.

209 Medium

209 of 248

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}

210 Medium

210 of 248

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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