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# NEET Chemistry Solved Question Paper 2010

#### Multiple Choice Questions

11.

Identify [A] and [B] in the following

${}_{89}{}^{227}\mathrm{Ac}\stackrel{-\mathrm{\beta }}{\to }\left[\mathrm{A}\right]\stackrel{-\mathrm{\alpha }}{\to }\left[\mathrm{B}\right]\stackrel{-\mathrm{\alpha }}{\to }\mathrm{Rn}$

• Po, Rn

• Th, Po

• Ra, Th

• Th, Ra

D.

Th, Ra

[A] and [B] in the above given reaction are Thorium (Th) and Radium (Ra).

${}_{89}{}^{227}\mathrm{Ac}\stackrel{-\mathrm{\beta }}{\to }{}_{90}{}^{227}\mathrm{Th}\stackrel{-\mathrm{\alpha }}{\to }{}_{88}{}^{223}\mathrm{Ra}$

12.

In haemoglobin, the metal ion present is

• Fe2+

• Zn2+

• Co2+

• Cu2+

A.

Fe2+

The metal ion present in haemoglobin is Fe2+.

13.

The bond order of CO molecule is

• 2

• 2.5

• 3

• 3.5

C.

3

The bond order of CO molecule is 3. It can be calculated as

CO ( 6 + 8 = 14)σ(1s)2,σ*(1s)2,σ(2s)2,σ*(2s)2,$\mathrm{\pi }$(2px)2≈σ(2pz)2

Bond Order =$\frac{{\mathrm{N}}_{\mathrm{b}}-{\mathrm{N}}_{\mathrm{a}}}{2}=\frac{10-4}{2}=3$

14.

Which of the following orbitals will have zero probability of finding the electron in the yZplane.

• px

• py

• pz

• dyz

A.

px

pxorbital lies along the x-axis only.

15.

Which of the following compounds shows evidence of the strongest hydrogen bonding?

• Propan-1-ol

• Propan-2-ol

• Propan-1,2-diol

• Propan-1,2,3-triol

D.

Propan-1,2,3-triol

Among all the given options, Propan-1, 2, 3 triol have strongest Hydrogen bonding as it has 3 polar - OH groups.

# 16.In Periodic Table, the basic character of oxidesincreases from left to right and decreases from top to bottom decreases from right to left and increases from top to bottom decreases from left to right and increases from top to bottom decreases from left to right and increases from bottom to top

C.

decreases from left to right and increases from top to bottom

On moving along a period, the tendency to lose electrons decreases and thus, the metallic character decreases. Since, the oxides of non-metals are acidic. Thus it can be said, along a period from left to right basic character decreases while on moving downward, basic character increases due to increase in metallic character.

17.

Which one of the following contains P-O-P bond?

• Hypophosphorus acid

• Phosphorus acid

• Pyrophosphoric acid

• Orthophosphoric acid

C.

Pyrophosphoric acid

Among all the given options, the correct option is Pyrophosphoric acid. The structures of the given acids is

18.

A weak acid of dissociation constant 10-5 is being titrated with aqueous NaOH solution. The pH at the point of one-third neutralisation of the acid will be

• 5 + log 2 - log 3

• 5 - log 2

• 5 - log 3

• 5 - log 6

B.

5 - log 2

The pH at the point of one-third neutralisation of the acid will be calculated as

Ka = 10-5 ; pKa = -log Ka = -log 10-5 = 5

$\underset{\mathrm{Final}}{\underset{\mathrm{Initial}}{}}\underset{=\frac{2}{3}\mathrm{mol}}{\underset{\left(1-\frac{1}{3}\right)\mathrm{mol}}{\underset{1\mathrm{mole}}{\mathrm{HA}}}}+\underset{0}{\mathrm{NaOH}}\to \underset{\frac{1}{3}\mathrm{mol}}{\underset{\frac{1}{3}\mathrm{mol}}{\underset{0}{\mathrm{NaA}}}}+\underset{\frac{1}{3}\mathrm{mol}}{\underset{0}{{\mathrm{H}}_{2}\mathrm{O}}}$

Here, final solution acts as an acidic buffer.

pH = pKa + log$\frac{\left[\mathrm{salt}\right]}{\left[\mathrm{acid}\right]}$

or, pH = 5 + log$\frac{\frac{1}{3}}{\frac{2}{3}}$

= 5 + log$\frac{1}{2}$

$\therefore$pH = 5 - log 2

19.

What type of orbital hybridisation is considered on P in PCl5?

• sp3d

• dsp3

• sp3d2

• d2sp3

A.

sp3d

In PCl5, the number of hybrid orbitals,

H =$\frac{1}{2}$[V + X + A - C]

where, V = number of valence electrons; X = number of monovalent atoms; A and C = charge of cation or anion]

H=$\frac{1}{2}$[5 + 5 + 0 - 0]

= 5

Therefore, the hybridisation is sp3d.

20.

Which of the following orders regarding ionisation energy is correct?

• N > O > F

• N < O < F

• N > O < F

• N < O < F

C.

N > O < F

Ionisation energy is the energy required to remove an electron from outer shell of an atom. Since, on moving from left to right in a period, size decreases, thus, larger amount of energy is required to remove an electron, i.e.ionisation energy increases.

The size of Oxygen atom is smaller as compared to Nitrogen but the ionisation energy of Nitrogen is higher. This is mainly because in Nitrogen, the electron is to be removed from half-filled 2p3 orbital.

Thus, the correct order of ionisation energy is O < N < F.