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# NEET Chemistry Solved Question Paper 2011

#### Multiple Choice Questions

1.

Which one of the following complexes is outer orbital complex?

• [Co(NH3)6]3+

• [Mn(CN)6]4+

• [Fe(CN)6]4-

• [Ni(NH3)6]2+

D.

[Ni(NH3)6]2+

Generally weaker field ligands form outer orbital complex. Between [Co(NH3)6]3+ and [Ni(NH3)6]2+, the later form outer orbital complex because of its d8configuration. (The configuration of Co3+ = d6).

2.

0.1 mol HCl is equal to

• 3.65 g

• 36.5 g

• 18 g

• 1.8 g

A.

3.65 g

Atomic weight of Chlorine = 35.5 gm

Atomic weight of Hydrogen = 1

Molecular mass of HCl = (35.5 + 1) gm = 36.5 gm

As 1 mole is the amount of the substance which has mass equal to gram molecular mass. Therefore, 1 mol HCl is equal to 36.5 gm and 0.1 mol HCl is equal to 3.65 gm.

3.

The geometry of XeF6 is

• planar hexagon

• regular octahedron

• distorted octahedron

• square bipyramid

C.

distorted octahedron

The geometry of XeF6 is distorted octahedral in which all the six positions are occupied by fluorine atoms and the lone pair of electrons of Xe atom is present at the comer of one of the triangular faces.

4.

10-6 M NaOH is diluted 100 times. The pH of the diluted base is

• between 7 and 8

• between 5 and 6

• between 6 and 7

• between 10 and 11

A.

between 7 and 8

[OH-] in the diluted base =$\frac{{10}^{-6}}{{10}^{2}}$= 10-8

Total [OH-] = 10-8 + [OH-] of water

= (10-8 + 10-7) M

= 10-8 [1 + 10] M

= 11 x 10-8 M

pOH = -log (11 x 10-8)

= - log 11 + 8 log 10

= 6.9586

pH = 14 - 6.9586 = 7.0414

5.

For the reaction,

H2O(l)$⇌$H2O(g)

at 373 K and 1 atm pressure

• $∆$H = 0

• $∆$E = 0

• $∆$H = T$∆\mathrm{S}$

• $∆$H = $∆$E

C.

$∆$H = T$∆\mathrm{S}$

$∆\mathrm{G}=∆\mathrm{H}-\mathrm{T}∆\mathrm{S}$

For the reaction,

H2O(l)$⇌$H2O(g)

At equilibrium,$∆$G = 0

Therefore,$∆\mathrm{H}=\mathrm{T}∆\mathrm{S}$

6.

The correct statement with regard to${\mathrm{H}}_{2}^{+}$and${\mathrm{H}}_{2}^{-}$is

• both H${}_{2}^{+}$ and H${}_{2}^{-}$ are equally stable

• both H${}_{2}^{+}$ and H${}_{2}^{-}$ do not exist

• H${}_{2}^{-}$ is more stable than H${}_{2}^{+}$

• H${}_{2}^{+}$ is more stable than H${}_{2}^{-}$

D.

H${}_{2}^{+}$ is more stable than H${}_{2}^{-}$

The correct statement with regard to${\mathrm{H}}_{2}^{+}$and${\mathrm{H}}_{2}^{-}$is, former is more stable than later.

${\mathrm{H}}_{2}^{+}:\mathrm{\sigma }1{\mathrm{s}}^{1},\mathrm{\sigma }1{\mathrm{s}}^{°}\phantom{\rule{0ex}{0ex}}\mathrm{Bond}\mathrm{order}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\mathrm{H}}_{2}^{-}:\mathrm{\sigma }1{\mathrm{s}}^{2},\mathrm{\sigma }1{\mathrm{s}}^{1}\phantom{\rule{0ex}{0ex}}\mathrm{Bond}\mathrm{order}=\frac{2-1}{2}=\frac{1}{2}$

The bond order of H${}_{2}^{+}$ and H${}_{2}^{-}$ are same.It is due to the presence of one electron in the antibonding molecular orbital in H${}_{2}^{-}$.

7.

Hybridisation states of C in CH${}_{3}^{+}$ and C${\mathrm{H}}_{3}^{-}$ are

• sp2, sp3

• sp3, sp2

• sp, sp2

• sp2, sp

A.

sp2, sp3

In CH${}_{3}^{+}$= lp + bp = 0 + 3 = 3

So, the hybridisation is sp2.

In${\mathrm{CH}}_{3}^{-}$= lp + bp = 1+ 3 = 4

So, the hybridisation is sp3.

Therefore, the correct answer is a.

8.

An sp3-hybrid orbital contains

A.

Each sp3- hybrid orbital has 25% or$\frac{1}{4}$s- character and 75% or$\frac{3}{4}$p- character.

# 9.A solution made by dissolving 40 g NaOH in 1000 g of H2O is1 molar 1 normal 1 molal None of these

C.

1 molal

Number of moles NaOH (n) =$\frac{\mathrm{Weight}\mathrm{in}\mathrm{g}}{\mathrm{molar}\mathrm{mass}}$=$\frac{40}{40}$= 1

As molality of a solution is defined as the number of moles of the solute dissolved in 1000 g (1 kg) of the solvent, thus the given solution of NaOH is 1 molal solution.

10.

Which of the following has the maximum number of unpaired electrons?

• V3+

• Fe2+

• Mn2+

• Cu+

C.

Mn2+

The outer electronic configuration of the given ions is as