Question

Find the number of species from the following which has spin only magnetic moment value of 1.73 BM. Fe$^{2+}$, Cu$^{2+}$, Ni$^{2+}$, NO, NO$_2$, NO$_2^-$, Sc$^{3+}$, Ti$^{3+}$

Answer 1

4

Answer 2

To find the number of species with a spin-only magnetic moment of 1.73 BM, we first need to determine the number of unpaired electrons corresponding to this magnetic moment.

The spin-only magnetic moment ($\mu_s$) is given by the formula:

$\mu_s = \sqrt{n(n+2)}$ BM

where 'n' is the number of unpaired electrons.

Given $\mu_s = 1.73$ BM:

$1.73 = \sqrt{n(n+2)}$

Squaring both sides:

$(1.73)^2 \approx 3$

$3 = n(n+2)$

$n^2 + 2n - 3 = 0$

Factoring the quadratic equation:

$(n+3)(n-1) = 0$

Since the number of unpaired electrons (n) cannot be negative, we take the positive value:

$n = 1$

Therefore, we need to identify all species from the given list that have exactly one unpaired electron.

Let's analyze each species:

1. Fe$^{2+}$:

   • Electronic configuration of Fe: $[Ar] 3d^6 4s^2$
   • Electronic configuration of Fe$^{2+}$: $[Ar] 3d^6$
   • In the $3d^6$ configuration, according to Hund's rule, the electrons are filled as: $\uparrow\downarrow \uparrow \uparrow \uparrow \uparrow$.
   • Number of unpaired electrons (n) = 4. (Does not match)

2. Cu$^{2+}$:

   • Electronic configuration of Cu: $[Ar] 3d^{10} 4s^1$
   • Electronic configuration of Cu$^{2+}$: $[Ar] 3d^9$
   • In the $3d^9$ configuration: $\uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow$.
   • Number of unpaired electrons (n) = 1. (Matches)

3. Ni$^{2+}$:

   • Electronic configuration of Ni: $[Ar] 3d^8 4s^2$
   • Electronic configuration of Ni$^{2+}$: $[Ar] 3d^8$
   • In the $3d^8$ configuration: $\uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow \uparrow$.
   • Number of unpaired electrons (n) = 2. (Does not match)

4. NO:

   • Nitric oxide (NO) has a total of 15 electrons (7 from N + 8 from O).
   • Its molecular orbital configuration is: $(\sigma_{1s})^2 (\sigma_{1s}^*)^2 (\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\sigma_{2p_z})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2 (\pi_{2p_x}^*)^1$.
   • It has one unpaired electron in the $\pi^*$ antibonding molecular orbital.
   • Number of unpaired electrons (n) = 1. (Matches)

5. NO$_2$:

   • Nitrogen dioxide (NO$_2$) has a total of 17 valence electrons (5 from N + 2*6 from O).
   • It is an odd-electron molecule, which means it must have at least one unpaired electron. Its Lewis structure shows one unpaired electron on the nitrogen atom.
   • Number of unpaired electrons (n) = 1. (Matches)

6. NO$_2^-$:

   • Nitrite ion (NO$_2^-$) has a total of 18 valence electrons (5 from N + 2*6 from O + 1 for the charge).
   • Since it has an even number of electrons, all electrons are paired in its ground state.
   • Number of unpaired electrons (n) = 0. (Does not match)

7. Sc$^{3+}$:

   • Electronic configuration of Sc: $[Ar] 3d^1 4s^2$
   • Electronic configuration of Sc$^{3+}$: $[Ar]$
   • It has no d electrons and no unpaired electrons.
   • Number of unpaired electrons (n) = 0. (Does not match)

8. Ti$^{3+}$:

   • Electronic configuration of Ti: $[Ar] 3d^2 4s^2$
   • Electronic configuration of Ti$^{3+}$: $[Ar] 3d^1$
   • In the $3d^1$ configuration: $\uparrow$.
   • Number of unpaired electrons (n) = 1. (Matches)

The species that have exactly one unpaired electron are Cu$^{2+}$, NO, NO$_2$, and Ti$^{3+}$. There are 4 such species.