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The three elements X, Y and Z with electronic configuration shown below all form hydrides :

$Element Electronic configuration X 1 s^{2} 2 s^{2} 2 p^{2} Y 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1} Z 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{5}$

Which set of properties match correctly with properties of the hydrides of these elements :

Hydride of X Hydride of Y Hydride of Z

Colourless gas insoluble in Silver/grey solid reacts with $H_{2} O$ Colourless gas forms a strong

$H_{2} O$ to form an alkali acid in $H_{2} O$

Colourless liquid, no reaction Silver/grey solid forms $H_{2}$ and Ionic solid with formula $ZH$

with $H_{2} O$ $H_{2} O$

Non-polar compound reacts Silver/grey ionic solid with Forms when water is added

with ${Cl}_{2}$ in light formula ${YH}_{2}$ to phosphorus and element Z

Colourless gas which burns Silver/grey solid which reacts Colourless, corrosive liquid

with air violently with acids at STP

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Explanation

(a) $X = {CH}_{4}, Y = NaH, Z = HBr$

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Which of the following is not true about helium?

It has the lowest boiling point

It has the highest first ionization energy

It can diffuse through rubber and plastic material

It can form clathrate compounds

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Explanation

(d) Due to small size of $He$ , it escapes from interstitial spaces/voids of

molecular lattice of quinols.

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${SbF}_{5}$ reacts with ${XeF}_{4}$ to form an adduct. The shapes of cation and anion in the adduct are respectively:

square planar, trigonal bipyramidal

T-shaped, Octahedral

square pyramidal, octahedral

square planar, octahedral

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Explanation

(b)

${XeF}_{4} + {SbF}_{5} \overset{}{\to} {[{XeF}_{3}]}^{+} {[{SbF}_{6}]}^{-} \overset{}{\to} \underset{\overset{{sp}^{3} d}{bent T shape}}{{[{XeF}_{3}]}^{+}} + \underset{\overset{{sp}^{3} d^{2}}{octahedral}}{{[{SbF}_{6}]}^{-}}$

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Select the correct statement about elements of group $15^{t h}$ .

The order of stability of oxidation state for +3 is $B i^{3 +} > S b^{3 +} > A s^{3 +}$ and for +5 is $B i^{5 +} < S b^{5 +} < A s^{5 +}$

In the case of nitrogen, all oxidation states from +1 to +4 tend to disproportionate in acid solution.

There is considerable increase in covalent radius from N to P but from As to Bi only a small increase in covalent radius is observed.

All of these.

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Explanation

(D) (A) This is due to the inert pair effect in the heavier p-block elements.

(B) $3 H N O_{2} \to H N O_{3} + H_{2} O + 2 N O$

(C) This is due to the presence of completely filled d and l or f orbitals in heavier elements. Covalent radius l pm; N =70, P =110, As =120, Bi =150.

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Assertion : $P_{4} O_{10}$ is not used to dry $N H_{3}$ .

Reason : $P_{4} O_{10}$ is acidic and $N H_{3}$ is basic.

A. If both the assertion and the reason are true and the reason is a correct explanation of the assertion B. If both the assertion and reason are true but the reason is not a correct explanation of the assertion C. If the assertion is true but the reason is false D. If both the assertion and reason are false

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Explanation

(A) $P_{4} O_{10} + 12 N H_{3} + 6 H_{2} O \to 4 {(N H_{4})}_{3} P O_{4}$

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Assertion : $H N O_{3}$ is stronger acid than $H N O_{2}$ .

Reason : In $H N O_{3}$ there are two nitrogen to oxygen bonds where as in $H N O_{2}$ there is only one.

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Explanation

(C)Conjugate base of $H N O_{3}$ , the $N O_{3}^{-}$ is more stable (charge dispersion on three oxygen atoms) than conjugate base of $H N O_{2}$ , the $N O_{2}^{-}$ (charge dispersion on two oxygen atoms).

There are three nitrogen to oxygen bonds in $H N O_{3}$ and two in $H N O_{2} .$

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The hybridisation of atomic orbitals of nitrogen in $N O_{2}^{+}, N O_{3}^{-} a n d N H_{4}^{+}$ are

sp, sp³ and sp² respectively

sp, sp² and sp³ respectively

sp², sp and sp³ respectively

sp², sp³ and sp respectively

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Explanation

No. of hybrid orbitals = Total no. of sigma bonds +lone pair of electrons. The no. of sigma bonds in $N O_{2}^{+}, N O_{3}^{-} a n d N H_{4}^{+}$ are 2,3 and 4 respectively. So the hybridisation is sp, sp² and sp³ respectively.

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Which of the following hybrides is most acidic

H₂Te

H₂Se

H₂O

H₂S

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Explanation

Due to weak bond H₂Te can furnish hydrogen more readily.The increasing order of acidic nature is

H₂O < H₂S < H₂Se < H₂Te

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Which of the following is incorrect about H₂SO₄

It can act as oxidising agent with copper

It acts as dehydrating agent with sugar

It acts as acid in its reaction with BaO₂

It acts as reducing agent in its reaction with Cu₂O

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Explanation

H₂SO₄ cannot act as reducing agent, because when a substance acts as reducing agent, it is oxidised itself and there is increase in O.N. In H₂SO₄the O.N. of sulphur is +6,which is maximum.

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Which of the following oxide is most acidic

Na₂O

Al₂O₃

P₂O₅

SO₃

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Explanation

The acidity decrease in the order SO₃> P₂O₅> Al₂O₃> Na₂O. The Al₂O₃is atmospheric, Na₂O is basic.

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