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NEET Practice Questions (MCQs) with Answers & Solutions

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In case of hydride of oxygen family, which of the following physical property change regularly on moving down the group.

Melting point

Thermal stability

Boiling point

Critical temperature

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Explanation

(b) Order of M.P. or B.P. or critical temperature : $H_{2} O > H_{2} Te > H_{2} Se > H_{2} S$

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In which case, the order of acidic strength is not correct ?

$HI > HBr > HCl$

${HIO}_{4} > {HBrO}_{4} > {HCIO}_{4}$

${HClO}_{4} > {HClO}_{3} > {HClO}_{2}$

$HF > H_{2} O > {NH}_{3}$

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Explanation

(b) ${HIO}_{4} > {HBrO}_{4} > {HCIO}_{4}$ $\Rightarrow$ acidic strength has been decided on the basis of

electronnegativity or charge density on central atom.

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Concentrated nitric acid reacts with iodine to give :

$HOI$

$HI$

${HIO}_{3}$

${HIO}_{4}$

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Explanation

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Thermally most stable compound is :

${HOCIO}_{3}$

${HOCIO}_{2}$

$HOCl$

$HOCIO$

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Explanation

(a) Decreasing order of thermal stability of oxy acids of chlorine.

${HOCIO}_{3}; {HOCIO}_{2}; HOCIO; HOCI$

In ${HOCIO}_{3}$ , chlorine is in +7 oxidation state.

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Which of the following halogen oxide is used for estimation of carbon monoxide in automobile exhaust gases?

${Cl}_{2} O_{7}$

$I_{2} O_{5}$

${ClO}_{2}$

${BrO}_{3}$

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Explanation

(b) $5 CO + I_{2} O_{5} \overset{}{\to} 5 {CO}_{2} + I_{2} I_{2} + 2 {Na}_{2} S_{2} O_{3} \overset{}{\to} 2 NaI + {Na}_{2} S_{4} O_{6}$

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Only iodine forms hepta-fluriode ${IF}_{7}$ , but chlorine and bromine give penta-flurides. The reason for this is :

low electron affinity of iodine

unusual pentagonal bipyramidal structure of ${IF}_{7}$

that the larger iodine atom can accommodate more number iof smaller fluorine atom around it

low chemical reactivity of ${IF}_{7}$

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Explanation

fluorine atoms around, it while due to smaller sizes of chlorine and

bromine atoms do not accommodate seven fluorine atoms, i.e., steric

factor dominate in case of chlorine and bromine.

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Which of the following has been arranged in order of decreasing bond dissociation energy :

$P — O > Cl — O > S — O$

$P — O > S — O > Cl — O$

$S — O > Cl — O > P — O$

$Cl — O > S — O > P — O$

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Explanation

(d) Bond dissociation energy $\propto \frac{1}{Bond length}$

Orders of bond length $P — O > S — O > Cl — O$ Order of $BDE :$ $Cl — O > S — O > P — O$

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Nitrogen gas is liberated by thermal decomposition of :

${NH}_{4} {NO}_{2}$

${NaN}_{3}$

${({NH}_{4})}_{2} {Cr}_{2} O_{7}$

All

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Explanation

(d)

$[\overset{‐ III}{{NH}_{4}} \overset{+ III}{{NO}_{2}}] \overset{∆}{\to} N_{2} ↑_{}^{} + 2 H_{2} O ↑_{}^{} {NaN}_{3} \overset{∆}{\to} Na + \frac{3}{2} N_{2} ↑_{}^{} {({NH}_{4})}_{2} {Cr}_{2} O_{7} \overset{}{\to} N_{2} ↑_{}^{} + \underset{(green)}{{Cr}_{2} O_{3} (s)} + 4 H_{2} O ↑_{}^{}$

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Two oxides of nitrogen, $NO$ and ${NO}_{2}$ are allowed to react together at $243 ° K$ and form a coloured compound of nitrogen (X). When compound (X) reacts with water to yield another compound of nitrogen (Y). The shape of the anion of (Y) molecule is :

triangular planar

triangle pyramidal

tetrahedron

square planar

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Among the following statement which one is true?

${NH}_{3}$ is less soluble than ${PH}_{3}$ in water

${NH}_{3}$ is stronger base and stronger reducing agent than ${PH}_{3}$

${NH}_{3}$ has higher boiling point than ${PH}_{3}$ and has lower melting point than ${PH}_{3}$

${PH}_{3}$ is stronger reducing agent than ${NH}_{3}$ and it has lower melting critical

temperature than ${NH}_{3}$

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Explanation

(d) ${NH}_{3}$ is a weak reducing agent than ${PH}_{3}$ , because $X — H$ bond strength

decreases down the group. Due to absence of H-bonding, only weak van

der waals force of attraction exists in ${PH}_{3}$ , it has lower critical

temperature than ${NH}_{3}$ .

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