Question

Peroxide ion _____
(i) Has five completely filled antibonding molecular orbitals.
(ii) Is diamagnetic
(iii)Has bond order one.
 (iv)Is isoelectronic with neon.

Which one of these is correct?
(ii) and (iii)
(i), (ii) and (iv)
(i), (ii) and (iii)
(ii) and (iv)

Answer 1

Hint:To solve this we must know the formula for the peroxide ion. The peroxide ion is . We have to find the number of antibonding molecular orbitals. The magnetic nature of peroxide ion can be determined from its electronic configuration. The number of covalent bonds in any molecule is known as its bond order. The species having the same number of electrons are known as isoelectronic.

Formula Used:
${\text{Bond order}} = \dfrac{1}{2}\left( {{\text{Number of electrons in bonding MO}} - {\text{Number of electrons in anti - bonding MO }}} \right)$

Complete step by step answer:1-0The total number of electrons in peroxide ions are 18. Thus, the electronic configuration of the peroxide ion is as follows:
${\text{O}}_2^{2 - }:\sigma _{1s}^2\sigma _{1s}^{*2}\sigma _{2s}^2\sigma _{2s}^{*2}\sigma _{2{p_z}}^2\pi _{2{p_x}}^2\pi _{2{p_y}}^2\pi _{2{p_x}}^{*2}\pi _{2{p_y}}^{*2}$
Thus, peroxide ion has four completely filled antibonding orbitals.
Thus, the statement ‘peroxide ion has five completely filled antibonding molecular orbitals’ is not correct.
2-From the electronic configuration, we can determine if peroxide ion is diamagnetic or not.
The total number of electrons in peroxide ions is 18. Thus, the electronic configuration of the peroxide ion is as follows:
${\text{O}}_2^{2 - }:\sigma _{1s}^2\sigma _{1s}^{*2}\sigma _{2s}^2\sigma _{2s}^{*2}\sigma _{2{p_z}}^2\pi _{2{p_x}}^2\pi _{2{p_y}}^2\pi _{2{p_x}}^{*2}\pi _{2{p_y}}^{*2}$
From the electronic configuration, we can see that there are no unpaired electrons i.e. all the electrons are paired. Thus, peroxide ion is diamagnetic.
Thus, the statement ‘peroxide ion is diamagnetic’ is correct.
3-We know that the formula to calculate the bond order is as follows:
${\text{Bond order}} = \dfrac{1}{2}\left( {{\text{Number of electrons in bonding MO}} - {\text{Number of electrons in anti - bonding MO }}} \right)$
From the electronic configuration, we can see that the number of electrons in bonding MO are 10 and the number of electrons in anti-bonding MO are 8. Thus,
${\text{Bond order}} = \dfrac{1}{2}\left( {10 - 8} \right) = 1$
Thus, the peroxide ion has bond order 1.
Thus, the statement ‘peroxide ion has bond order 1’ is correct.
4-We know that the species having the same number of electrons are said to be isoelectronic.
The total electrons of peroxide ions are 18. The atomic number of neon is 10 and thus, the total number of electrons of neon are 10.
Thus, peroxide ion is not isoelectronic with neon.
Thus, the statement ‘peroxide ion is isoelectronic with neon’ is not correct.
Thus, the correct statements are (ii) and (iii).

Thus, the correct option is (A).

 Note:To solve this remember the formula to calculate the bond order. The atomic number of argon is 18 and thus, the number of electrons of argon are 18. Thus, peroxide ion is isoelectronic with argon and not with neon.