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# NEET Chemistry Solved Question Paper 2010

#### Multiple Choice Questions

1.

Which of the following has the highest bond order?

• N2

• O2

• He2

• H2

A.

N2

N2( 7 + 7 = 14e-) = KK*$\mathrm{\sigma }$2s2$\stackrel{*}{\mathrm{\sigma }}$2s2$\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^{2}$

BO    = $\frac{1}{2}$(8 - 2) = 3

O(8 + 8 =16e-) = KK*$\mathrm{\sigma }$2s2$\stackrel{*}{\mathrm{\sigma }}$2s2$\mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^{2}$

BO   =

He( 2 + 2 = 4e-) = $\sigma$1s2$\mathrm{\sigma }$*1s2

BO = $\frac{1}{2}$(2 - 2) = 0

H2 (1 + 1 = 2e-) = $\mathrm{\sigma }$1s2

BO   = $\frac{1}{2}$(2 - 0) = 1

Hence, N2 has the highest bond order.

2.

The quantum numbers +$\frac{1}{2}$ and -$\frac{1}{2}$ for the electron spin represent

• rotation of the electron in clockwise and anticlockwise direction respectively

• rotation of the electron in anticlockwise and clockwise direction respectively

• magnetic moment of the electron pointing up and down respectively

• two quantum mechanical spin states which have no classical analogues

A.

rotation of the electron in clockwise and anticlockwise direction respectively

The quantum numbers + $\frac{1}{2}$ and - $\frac{1}{2}$ indicate the rotation of electrons around its axis in opposite directions. Hence, they represent the two quantum mechanical spin states which have no classical analogues.

3.

Which of the following relation is correct ?

• Ist IE of C > Ist IE of B

• Ist IE of C < Ist IE of B

• IInd IE of C > IInd IE of B

• Both (b) and (c)

A.

Ist IE of C > Ist IE of B

The minimum amount of energy required to remove the most loosely bound electron from an isolated gaseous atom so as to convert it into a gaseous cation, is called its ionisation energy. Because the size of C is smaller than that of the B due to effective nuclear charge, hence

Ist IE of C > Ist IE of B

4.

The density of a gas is found to be 1.56 g/L at 745 mm pressure and 65°C. What is the moleculer mass of the gas?

• 44.2 u

• 4.42 u

• 2.24 u

• 22.4 u

A.

44.2 u

Given, p = 745 mm =

T = 65$°$C = 65 + 273 = 338 K

5.

Considering H2O as weak field ligand, the number of unpaired electrons in [Mn(H2O)6] will be (Atomic number of Mn = 25)

• five

• two

• four

• three

A.

five

Mn in [Mn(H2O)6]2+ is present as Mn2+(3d5).

sp3d2 hybridisation occurs in[Mn(H2O)6]+2. Due to the presence of weak field ligand pairing of electron does not take place, hence the number of unpaired electrons is five.

6.

The correct order towards bond angle is

• sp3 < sp2 < sp

• sp < sp2 < sp3

• sp < sp3 < sp2

• sp2 < sp3 < sp

A.

sp3 < sp2 < sp

Bond angle is defined as the angle between the orbitals containing bonding electron pairs around the central atom in a  molecule/complex ion.

s character $\propto$ bond angle

Hence, the correct order is sp3 < sp2 < sp

7.

Which of the following configurations corresponds to element of highest ionisation energy?

• 1s2, 2s1

• 1s2, 2s2, 2p3

• 1s2, 2s2, 2p2

• 1s2, 2s2, 2p6, 3s1

B.

1s2, 2s2, 2p3

The ionisation energy of 1s2, 2s2, 2p3 configuration is highest due to presence of half- filled orbitals.

8.

In long form of Periodic Table, the properties of the elements are a periodic function of their

• atomic size

• ionisation energy

• atomic mass

• atomic number

D.

atomic number

In modern Periodic Table, the physical and chemical properties of the elements are a periodic function of their atomic number, i.e., if the elements are arranged in order of their increasing atomic number, the elements with similar properties are repeated after certain regular intervals.

9.

In the electronic structure of H2SO4 , the total number of unshared electrons is

• 20

• 16

• 12

• 8

A.

20

The structure of H2SO4 is

Total number of unshared electrons = 10 x 2 = 20

Therefore, among all the given options, option a is correct.

10.

Which of the following has a bond formed by overlap of sp-sp3 hybrid orbitals?

• CH3 - C $\equiv$ C - H

• CH3 - CH = CH - CH3

• CH2 = CH - CH = CH2

• HC $\equiv$CH

A.

CH3 - C $\equiv$ C - H

CH3 - C $\equiv$ C - H has a bond formed by overlap of sp-sp3 hybrid orbitals. Other options consist of following orbitals:

b. CH3 - CH = CH - CH3

It has a bond overlapped by sp3-sp2-sp2-sp3 orbital.

c. CH2 = CH - CH = CH2

It has a bond overlapped by sp2 orbital.

d. HC $\equiv$CH

It has a bond overlapped by sp orbital.