Periodic Table & Periodicity — Page 4 | JEE Chemistry

Q31

The correct order of bond dissociation enthalpy of halogens is :

A F2 > Cl2 > Br2 > I2

B Cl2 > Br2 > F2 > I2

C I2 > Br2 > Cl2 > F2

D Cl2 > F2 > Br2 > I2

Correct Answer

Option B

Solution

Among halogens (F2, Cl2, Br2 and I2), bond dissociation enthalpy (

\Delta

dissH

^\circ

) of I2, is minimum because of larger size of I-atom there is a steric repulsion between bonded I-atoms, which makes I-I bond weakest.

Whereas, smaller size and highest electronegativity of F-atom cause highest electron density on F-atom of F2 molecule.

As a result, F-F bond becomes weaker due to electrostatic repulsion between bonded F-atoms.

Thus, the order of

\Delta

dissH

^\circ

(in kJ mol $-$ 1) is

\mathop {Cl - Cl}\limits_{(242.6)} > \mathop {Br - Br}\limits_{(192.3)} > \mathop {F - F}\limits_{(158.8)\,Electrostatic\,repulsion} > \mathop {I - I}\limits_{(151.1)\,Steric\,repulsion}

Q32

The correct order of hydration enthapies of alkali metal ions is :

A Li+ > Na+ > Cs+ > Rb+

B Na+ > Li+ > K+ > Cs+ > Rb+

C Na+ > Li+ > K+ > Rb+ > Cs+

D Li+ > Na+ > K+ > Rb+ > Cs+

Correct Answer

Option D

Solution

Hydration energy/enthalpy $\propto$

{{ch\arg e} \over {size}}

$\therefore$ Smaller the size more is hydration energy.

Li+ having minimum radius so maximum hydration energy.

Correct order is : Li+ > Na+ > K+ > Rb+ > Cs+

Q33

Portland cement contains 'X' to enhance the setting time. What is 'X'?

A

\mathrm{CaSO}_{4} \cdot \dfrac{1}{2} \mathrm{H}_{2} \mathrm{O}

B

\mathrm{CaSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}

C

\mathrm{CaSO}_{4}

D

\mathrm{CaCO}_{3}

Correct Answer

Option B

Solution

Gypsum is added in portland cement to slow down the process of setting of the cement so that it gets sufficient time to hardened.

Q34

The element having greatest difference between its first and second ionization energies, is :

A Ba

B Ca

C Sc

D K

Correct Answer

Option D

Solution

Electronic configuration of Potassium (K) : 1s22s22p63s23p64s1 After first ionisation enthalpy its configuration becomes : 1s22s22p63s23p6 which is a inert gas configuration.

To remove an electron from inert gas configuration requires very high energy.

So Potassium (K) have high difference in the first ionisation and the second ionisation energy as it achieve stable noble gas configuration after first ionisation.

Q35

The lathanide ion that would show colour is :

A Sm3+

B Gd3+

C Lu3+

D La3+

Correct Answer

Option A

Solution

Electronic configuration of Sm = [Xe] 4f6 6s2 $\therefore$ Electronic configuration of Sm+3 = [Xe] 4f5 Sm+3 shows yellow colour due to partially filled f orbital and f-f transition.

Electronic configuration of La+3 = [Xe] 4f0 Lu+3 = [Xe] 4f14 Gd+3 = [Xe] 4f7 La+3, Lu+3 and Gd+3 are colourless due to absence of f-f transition.

Q36

C{e^{ + 3}},\,\,L{a^{ + 3}},\,\,P{m^{ + 3}}\,\,

and

Y{b^{ + 3}}\,\,

have ionic radial in the increasing order as

A

L{a^{ + 3}}\,\, < C{e^{ + 3}}\,\, < P{m^{ + 3}}\,\, < Y{b^{ + 3}}

B

Y{b^{ + 3}}\,\, < P{m^{ + 3}}\,\, < C{e^{ + 3}}\,<\,L{a^{ + 3}}

C

L{a^{ + 3}}\,\, = C{e^{ + 3}}\,\, < P{m^{ + 3}}\,\, < Y{b^{ + 3}}

D

Y{b^{ + 3}}\,\, < P{m^{ + 3}}\,\, < L{a^{ + 3}}\,\, < C{e^{ + 3}}

Correct Answer

Option B

Solution

Here

C{e^{ + 3}},\,\,P{m^{ + 3}},\,\,Y{b^{ + 3}}

are from lanthanide series.

L{a^{ + 3}}

atomic number is

57.

This is a

d

block element but all the property of

La

is similar to lanthanides. It is present before

Ce.

As we know in lanthanides the radius of elements decreases from left to right of periodic table. So, order will be

L{a^{ + 3}} > C{e^{ + 3}} > P{m^{ + 3}} > Y{b^{ + 3}}

Q37

The correct order of ionic radius is

A

Ce > Sm > Tb > Lu

B

Lu > Tb > Sm > Ce

C

Tb > Lu > Sm > Ce

D

Sm > Tb > Lu > Ce

Correct Answer

Option A

Solution

Those are the elements of lanthanides series. In lanthanide series the extra electron is added in

4f

subshell and we know in periodic table from left to right the effective nuclear charge increases as electron is added so the size of elements will decreases.

So, the correct order is

Ce > Sm > Tb > Lu.

Q38

The atomic numbers of Vanadium (V), Chromium (cr), Manganese (Mn) and Iron (Fe), respectively,

23,24,25

and

26

. Which one of these may be expected to have the higher second ionization enthalpy?

A Cr

B Mn

C Fe

D V

Correct Answer

Option A

Solution

Ionization enthalpy is the energy required to remove the electron from the outer most orbit. Electronic configuration :

\begin{aligned}& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,V\left( {23} \right) = \left[ {Ar} \right]\,\,\,3{d^3}\,\,\,4{S^2} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,Cr\left( {24} \right) = \left[ {Ar} \right]\,\,\,3{d^5}\,\,\,4{S^1} \\ & \,\,\,\,\,\,\,\,\,\,\,Mn\left( {25} \right) = \left[ {Ar} \right]\,\,\,3{d^5}\,\,\,4{S^2} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,Fe\left( {26} \right) = \left[ {Ar} \right]\,\,\,3{d^6}\,\,\,4{S^2}\end{aligned}

After first ionization enthalpy the electronic configuration will be,

\begin{aligned}& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,V\left( {23} \right) = \left[ {Ar} \right]\,\,\,3{d^3}\,\,\,4{S^1} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,Cr\left( {24} \right) = \left[ {Ar} \right]\,\,\,3{d^5} \\ & \,\,\,\,\,\,\,\,\,\,\,Mn\left( {25} \right) = \left[ {Ar} \right]\,\,\,3{d^5}\,\,\,4{S^1} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,Fe\left( {26} \right) = \left[ {Ar} \right]\,\,\,3{d^6}\,\,\,4{S^1}\end{aligned}

Now in 2nd ionization enthalpy one more electron will be removed.

And after second ionization enthalpy electronic configuration will be,

\begin{aligned}& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,V\left( {23} \right) = \left[ {Ar} \right]\,\,\,3{d^3} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,Cr\left( {24} \right) = \left[ {Ar} \right]\,\,\,3{d^4} \\ & \,\,\,\,\,\,\,\,\,\,\,Mn\left( {25} \right) = \left[ {Ar} \right]\,\,\,3{d^5} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,Fe\left( {26} \right) = \left[ {Ar} \right]\,\,\,3{d^6}\end{aligned}

For

V(23), Mn (25)

and

Fe (26)

electron is removed from

4S

orbital but for

Cr (24)

electron is removed from

3d

orbital. We know distance of

4S\,\, > \,\,3d

from the nucleus. So, the attraction on the electrons of

45

shell is less compared to

3d

shell by the nucleus. So, the removed of electron will be easier for the electrons of

4S

shell than

3d

shell. So, for

Cr(24)

second ionization is more than other as electron is removed from

3d

shell. And for

Cr(24)

outer most shell

''3d''

was half filled after

1st

ionization enthalpy, so

3d

shell was stable. So removal of electron will be difficult from stable shell. That is why also

2nd

ionization enthalpy of

Cr(24)

is higher compare to other given atoms.

Q39

Which one of the following is an amphoteric oxide?

A Na2O

B SO2

C B2O3

D ZnO

Correct Answer

Option D

Solution

These elements are called amphoteric which reacts with both acid and base. Following elements are amphoteric

\left( 1 \right)\,\,\,\,Zn

\left( 2 \right)\,\,\,\,Be

\left( 3 \right)\,\,\,\,Al

\left( 4 \right)\,\,\,\,B

\left( 5 \right)\,\,\,\,Cr

\left( 6 \right)\,\,\,\,Ga

\left( 7 \right)\,\,\,\,Pb

\left( 8 \right)\,\,\,\,Sn

Here

Zn,Be,Al,Pb\,\,\,

and

\,

Sn

are mostly amphoteric.

B, Cr

and

Ga

are less amphoteric.

Q40

The reduction in atomic size with increase in atomic number is a characteristic of elements of

A

d

- block

B

f

- block

C Radioactive series

D High atomic masses.

Correct Answer

Option D

Solution

This can't be

f

- block elements as

f

block means only lanthanides and actinides but we know lanthanide contraction happens on the elements before lantanidine elements also.

So, radius also decrease for those element when increase the atomic number.

Radioactive Series starts from atomic number

84

( Polonium ) and affer that all are radioactive. But we know before

84

atomic number also the atomic size decrease with increase in atomic number.

So, only radioactive series don't show this characteristic.

So, correct answer should be high atomic mass.