Question

The sum of number of unpaired electrons in each of the following species is

$B_2$, XeF, $[Fe(H_2O)_5(NO)]SO_4$, $K_3CrO_8$, $Ba(O_2)_2$

Answer 1

5

Answer 2

The sum of the number of unpaired electrons in each of the given species is calculated as follows:

1. $B_2$: Boron ($Z=5$) has electronic configuration $1s^2 2s^2 2p^1$. $B_2$ molecule has a total of $2 \times 3 = 6$ valence electrons. According to Molecular Orbital Theory, the order of filling MOs for $B_2$ is $\sigma_{2s} < \sigma^*_{2s} < \pi_{2p} < \sigma_{2p}$. The 6 valence electrons fill the MOs as $(\sigma_{2s})^2 (\sigma^*_{2s})^2 (\pi_{2p})^2$. The $\pi_{2p}$ molecular orbitals are degenerate. According to Hund's rule, the two electrons in the $\pi_{2p}$ orbitals occupy the two degenerate orbitals singly. Thus, there are 2 unpaired electrons in $B_2$.

2. XeF: Xenon ($Z=54$) has 8 valence electrons. Fluorine ($Z=9$) has 7 valence electrons. The XeF molecule has a total of $8 + 7 = 15$ valence electrons. With an odd number of electrons, XeF is a radical and must have at least one unpaired electron. The most common representation of XeF radical shows an unpaired electron on the Xe atom. Thus, there is 1 unpaired electron in XeF.

3. $[Fe(H_2O)_5(NO)]SO_4$: This is the brown ring complex. The salt is composed of $[Fe(H_2O)_5(NO)]^{2+}$ cation and $SO_4^{2-}$ anion. The sulfate anion ($SO_4^{2-}$) contains only paired electrons (S in +6 oxidation state, O in -2 oxidation state, all valence electrons involved in bonding or lone pairs). We need to find the number of unpaired electrons in the complex cation $[Fe(H_2O)_5(NO)]^{2+}$. In this complex, Fe is typically considered to be in the +1 oxidation state, and the NO ligand is considered as $NO^+$. The complex can be described as a {Fe(NO)}$^7$ system, where the superscript 7 is the sum of the d electrons on Fe ($3d^7$ for Fe$^+$) and the $\pi^*$ electrons on $NO^+$ (0 for $NO^+$). The 7 electrons in the {M(NO)}$^n$ system occupy the molecular orbitals formed from metal d orbitals and NO $\pi^*$ orbitals. The filling order is $(e_1)^4 (a_1)^2 (e_2)^1$. The $e_2$ level is doubly degenerate and contains 1 electron. Thus, there is 1 unpaired electron in $[Fe(H_2O)_5(NO)]^{2+}$.

4. $K_3CrO_8$: This is a salt containing the peroxochromate anion, $[Cr(O_2)_4]^{3-}$. The ligand is the peroxide ion, $O_2^{2-}$, which is bidentate. Let the oxidation state of Cr be $x$. Each peroxide ion has a charge of -2. So, $x + 4 \times (-2) = -3$, which gives $x = +5$. Chromium ($Z=24$) has electronic configuration $[Ar] 3d^5 4s^1$. $Cr^{5+}$ has electronic configuration $[Ar] 3d^1$. In any ligand field, a single d electron will occupy one orbital and remain unpaired. Thus, there is 1 unpaired electron in $K_3CrO_8$.

5. $Ba(O_2)_2$: Barium (Ba) is in Group 2, so it is in the +2 oxidation state ($Ba^{2+}$). The anion must have a total charge of -2 to balance the +2 charge of Ba. The formula $Ba(O_2)_2$ indicates that the anion is $(O_2)_2^{2-}$, which means there are two $O_2$ units with a total charge of -2. This corresponds to two peroxide ions, $O_2^{2-}$. So, the compound is Barium peroxide, $Ba(O_2)_2$, containing $Ba^{2+}$ and two $O_2^{2-}$ ions. $Ba^{2+}$ has the electronic configuration of Xe, which is $[Kr] 4d^{10} 5s^2 5p^6$, containing only paired electrons. The peroxide ion, $O_2^{2-}$, has $2 \times 6 + 2 = 14$ valence electrons. The MO configuration for $O_2^{2-}$ is $(\sigma_{2s})^2 (\sigma^*_{2s})^2 (\sigma_{2p_x})^2 (\pi_{2p})^4 (\pi^*_{2p})^4$. All electrons are paired. Thus, $O_2^{2-}$ has 0 unpaired electrons. Since both $Ba^{2+}$ and $O_2^{2-}$ have 0 unpaired electrons, $Ba(O_2)_2$ has 0 unpaired electrons.

The number of unpaired electrons in each species are: $B_2$: 2 XeF: 1 $[Fe(H_2O)_5(NO)]SO_4$: 1 $K_3CrO_8$: 1 $Ba(O_2)_2$: 0

The sum of the number of unpaired electrons in each of the species is $2 + 1 + 1 + 1 + 0 = 5$.