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NEET Chemistry Solved Question Paper 2015

Multiple Choice Questions

11.

In which of the following pairs, both the species are not isostructural?

SiCl_4.PCl₄⁺
Diamond, carbide
NH₃, PH₃
NH₃, PH₃

NH₃, PH₃

a) SiCl_4.PCl₄⁺: Both are isostructural because their central atom is sp³ hybridised and both have a tetrahedral arrangement.

b) Diamond and silicon carbide (SiC): Both are isostructural because their central atom is sp³ hybridised and both have a tetrahedral arrangement.

c) NH₃ and PH₃ have sp³ geometry.

d) XeF₄ has sp³d² hybridizations while XeO₄ has sp³ hybridizations.Hence, XeF₄ and XeO₄ are not isostructural.

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12.

What is the mole fraction of the solute in a 1.00 m aqueous solution?

0.177
1.770
0.0354
0.0354

0.0354

Molality space left parenthesis straight m right parenthesis space equals space fraction numerator 1000 space straight x space straight n over denominator straight N space straight x space straight M end fraction where comma space straight n space equals space number space of space moles space of space solute straight N space equals space number space of space moles space of space solvent straight M space equals space molar space mass space of space solvent Given comma space straight m space equals space 1 1 space equals space fraction numerator 1000 space straight x space straight n over denominator straight N space straight x space 18 end fraction space rightwards double arrow space straight n over straight N space equals space 18 over 1000 Or space fraction numerator straight n over denominator straight n space plus space straight N end fraction space equals space 18 over 1018 space equals space 0.0177 Alternat space method space straight N space equals space 1000 over 18 space equals 55.5 space mol left square bracket therefore space 1 space straight m space solution space implies space that space 1 space mole space of space solute space is space present space in space 1 space kg space of space 1000 space straight g space water right square bracket therefore comma Mole space fraction space of space solute space equals space fraction numerator straight n over denominator straight n space plus straight N end fraction space equals space fraction numerator 1 over denominator 1 plus 55.5 end fraction space equals space 0.0177

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13.
Two possible stereo-structures of CH₃CHOH.COOH, which are optically active, are called

diastereomers

atropisomers

enantiomers

enantiomers

enantiomers

Both are enantiomers.

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14.

Which of the following statement is correct for a nucleophile?

Nucleophile is s Lewis acid
Ammonia is a nucleophile
Nucleophiles attack low electrons density sites
Nucleophiles attack low electrons density sites

Nucleophile is s Lewis acid

Nucleophiles are electron rich species. Hence, act as a Lewis base but not Lewis acid.

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15.

If the equilibrium constant for

$straight N subscript 2 space left parenthesis straight g right parenthesis space plus space straight O subscript 2 space left parenthesis straight g right parenthesis space rightwards harpoon over leftwards harpoon space 2 space NO space left parenthesis straight g right parenthesis space is space straight K comma space the space equilibrium space constant space for 1 half straight N subscript 2 space left parenthesis straight g right parenthesis space plus space 1 half straight O subscript 2 left parenthesis straight g right parenthesis space rightwards harpoon over leftwards harpoon space NO space left parenthesis straight g right parenthesis space will space be$

K^1/2
K/2
K
K

K^1/2

As we can see the reaction for which we have to find out equilibrium constant is different only in stoichiometric coefficient as compared to the given reaction. Hence, we can find equilibrium constant as,

$straight N subscript 2 space left parenthesis straight g right parenthesis space plus space straight O subscript 2 space left parenthesis straight g right parenthesis space rightwards harpoon over leftwards harpoon space 2 space NO space left parenthesis straight g right parenthesis thin space is space straight K straight i. straight e. space straight K space equals space fraction numerator left square bracket NO right square bracket squared over denominator left square bracket straight N subscript 2 right square bracket left square bracket straight O subscript 2 right square bracket end fraction space space space.. space left parenthesis straight i right parenthesis Let space equilibrium space constant space for space the space reaction comma 1 half space straight N subscript 2 space left parenthesis straight g right parenthesis space plus space 1 half straight O subscript 2 space left parenthesis straight g right parenthesis space rightwards harpoon over leftwards harpoon space 2 NO space left parenthesis straight g right parenthesis space is space straight K apostrophe straight i. straight e space space straight K apostrophe space equals space fraction numerator left square bracket NO right square bracket over denominator left square bracket straight N subscript 2 right square bracket to the power of 1 divided by 2 end exponent left square bracket straight O subscript 2 right square bracket to the power of 1 divided by 2 end exponent end fraction On space squaring space both space sides straight K apostrophe squared space equals space fraction numerator left square bracket NO right square bracket squared over denominator left square bracket straight N subscript 2 right square bracket left square bracket straight O subscript 2 right square bracket end fraction space space.. space left parenthesis ii right parenthesis On space comparing space Eqs. space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis space we space get straight K equals space straight K apostrophe squared or straight K to the power of apostrophe space equals space square root of straight K$

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16.

Gadolinium belongs to 4f series. Its atomic number is 64. which of the following is the correct electronic configuration of gadolinium?

[Xe]4f⁸ 6d²
[Xe]4f⁹5s¹
[Xe]4f⁷ 5d¹6s²
[Xe]4f⁷ 5d¹6s²

[Xe]4f⁷ 5d¹6s²

Atomic number of Gadolinium = 64

electronic configuration of Gd,

[Xe]4f⁷ 5d¹6s²

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17.

An aqueous solution of which of the following compounds is the best conductor of electric current?

Acetic acid C₂H₄O₂
Hydrochloric acid, HCl
Ammonia, NH₃
Ammonia, NH₃

Hydrochloric acid, HCl

HCl is strong acid and dissociates completely. Hence, it conducts electricity best in its aqueous solution.

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18.

On heating which of the following releases CO₂ most easily?

K₂CO₃
Na₂CO₃
MgCO₃
MgCO₃

MgCO₃

Order of thermal stability is

K₂CO₃ > Na₂CO₃ > CaCO₃ > MgCO₃Thus, MgCO₃ releases CO₂ most easily

MgCO₃ + heat ---> MgO + CO₂

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19.

The number of water molecules is maximum in

18 molecules of water
1.8 g of water
18 g of water
18 g of water

18 g of water

Mole is the biggest unit to measure the number of molecules/atom/ions.

therefore,

1 mole of water contains molecules = 6.02 x 10²³

therefore, 18 moles of water contain molecules

= 18 x 6.02 x 10²³ molecules

Now, 1 mole of water = 18 g of water

Hence, number of molecules in 18 g of water
= 6.02 x10²³

and 1.8 g of water contains = 6.02 x 10²² molecules.

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