Question

which of the following have an incorrect order of ionization energy
1.Pb>Sn
2.Na+>Mg+
3.Li+>O+
4.Be+<C+​

Answer 1

Explanation:

2.Na+>Mg+..........

Answer 2

Answer:

The pair showing the incorrect order of ionization energy is Be⁺ < C⁺.

Therefore, option 4) Be⁺ < C⁺ is correct.

Explanation:

The pair showing the incorrect order of ionization energy =?

Ionization energy: The energy required to remove an electron from the valence shell of the atom or an ion.

1) Pb > Sn

Here,

• Pb is lead ( atomic number 82).
• Sn is tin ( atomic number 50 ).

Pb and Sn both come in the same group. And as we know, by going down in the group, the ionization energy decreases as the atomic size increases.

But, in the case of Pb and Sn;

• The atomic size of Pb and Sn is very close due to lanthanoid contraction.
• In the case of Pb, there is more charge density than Sn, which causes more attraction of outer electrons toward the nucleus.
• Hence, the ionisation energy of Pb is greater than Sn.

Therefore, Pb > Sn ( correct )

2) Na⁺ > Mg⁺

Here,

• Na is sodium ( Z = 11 ) and Mg is magnesium ( Z = 12 )

In the case of Na⁺;

• Electronic configuration of Na = $1s^{2}2s^{2}2p^{6}3s^{1}$
• Electronic configuration of Na⁺ = $1s^{2}2s^{2}2p^{6}$
• It has a fully filled 2p orbital; thus, it will be difficult to remove an electron from Na⁺.

In the case of Mg⁺;

• Electronic configuration of Mg = $1s^{2}2s^{2}2p^{6}3s^{2}$
• Electronic configuration of Mg⁺ = $1s^{2}2s^{2}2p^{6}3s^{1}$
• It has a half-filled 3s which is away from the nucleus as compared to the 2p orbital; thus, it will be less difficult to remove an electron from Mg⁺ than Na⁺.

Hence, Na⁺ > Mg⁺ ( correct )

3) Li⁺ > O⁺

Here,

• Li is lithium ( Z = 3 ) and O is oxygen ( Z = 8 )

In the case of Li⁺;

• Electronic configuration of Li = $1s^{2}2s^{1}$
• Electronic configuration of Li⁺ = $1s^{2}$
• It has a fully filled 1s orbital; thus, it will be difficult to remove an electron from Li⁺.

In the case of O⁺;

• Electronic configuration of O = $1s^{2}2s^{2}2p^{4}$
• Electronic configuration of O⁺ = $1s^{2}2s^{2}2p^{3$
• It has a half-filled 2p orbital which is away from the 1s orbital; thus, it will be less difficult to remove an electron from O⁺ than Li⁺.

Hence, Li⁺ > O⁺ ( correct )

4) Be⁺ < C⁺

Here,

• Be is beryllium ( Z = 4 ) and C is carbon ( Z = 6 )

In the case of Be⁺;

• Electronic configuration of Be = $1s^{2}2s^{2}$
• Electronic configuration of Be⁺ = $1s^{2}2s^{1}$
• It has a half-filled 2s orbital which is very close to the nucleus; thus, it will be difficult to remove an electron from Be⁺.

In the case of C⁺;

• Electronic configuration of C = $1s^{2}2s^{2}2p^{2}$
• Electronic configuration of C⁺ = $1s^{2}2s^{2}2p^{1}$
• It has an unpaired electron in the p-orbital; thus, it will be less difficult to remove an electron from C⁺ than Be⁺.

Hence, Be⁺ < C⁺ ( incorrect ).