Mn Nitride Bonding Analysis

Below is a structured outline of how one typically arrives at each of the answers. Note that many of these points hinge on “formal” oxidation‐state assignments and standard MO arguments for a linear (or nearly linear) M≡N multiple bond in $C_{4v}$ symmetry.

(i) Oxidation State and $d$-Electron Count

We are told that the complex $\{PW_{11}O_{39}(MnN)\}^{5-}$ is formed upon loss of $N_2$ from $\{PW_{11}O_{39}(Mn-N_3)\}^{5-}$. The polyanion “framework” $\{PW_{11}O_{39}\}$ is usually assigned a charge of $-7$.

1. Azide complex

   • Overall charge: $-5$.
   • Polyoxometalate (POM) fragment: $-7$.
   • Azide ($\mathrm{N_3^-}$) has a formal charge of $-1$.
   • Let the metal’s oxidation state be $x$. Then $$(-7) + (-1) + x \;=\; -5 \;\;\Longrightarrow\;\; x=+3.$$
   • Hence in the azide species, the manganese is formally Mn(III).
   • Manganese’s ground‐state electron configuration is $[\mathrm{Ar}]\,3d^5\,4s^2$. Removing 3 electrons gives $d^4$.

2. Nitrido complex

   • Overall charge: $-5$.
   • Polyoxometalate: $-7$.
   • A terminal nitrido ($\mathrm{N}^{3-}$) is assigned $-3$.
   • Let the oxidation state of Mn be $y$. Then $$(-7)\;+\;(-3)\;+\;y\;=\;-5 \;\;\Longrightarrow\;\; y=+5.$$
   • Therefore, in the nitrido complex, manganese is formally Mn(V) and thus has $d^2$ configuration.

Most questions about the “bonding” in a high‐valent metal–nitrido generally focus on the final Mn(V), $d^2$ state.

(ii) Partial MO Diagram for a Tetragonal $\mathrm{M}\equiv\mathrm{N}$ Bond

Take the $z$-axis along the M–N bond and use $C_{4v}$ symmetry. The key points:

• The nitrido ($\mathrm{N}^{3-}$) has three 2p orbitals (p$_x$, p$_y$, p$_z$) relevant for bonding.
• In a linear (or nearly linear) arrangement, $\mathrm{p}_z$ will be the $\sigma$-type orbital; $\mathrm{p}_x$ and $\mathrm{p}_y$ can serve as $\pi$-type orbitals.
• The metal $d$-orbitals split in $C_{4v}$ (with the z‐axis as the principal axis) into the following irreps: $$d_{z^2} \;\to\; A_1,\quad d_{x^2-y^2} \;\to\; B_1,\quad d_{xy} \;\to\; B_2,\quad (d_{xz},\,d_{yz}) \;\to\; E.$$
• The nitrido orbital that is along $z$ ($\mathrm{p}_z$) also transforms as $A_1$. Hence it $\sigma$-couples with the metal $d_{z^2}$.
• The pair ($\mathrm{p}_x$, $\mathrm{p}_y$) transforms as $E$. They form $\pi$-type bonds with the metal $d_{xz},\, d_{yz}$ (also $E$).
• The remaining metal orbitals, $d_{x^2-y^2}(B_1)$ and $d_{xy}(B_2)$, have no partner orbitals on the nitrido of the same symmetry and thus end up nonbonding in this simple view.

A common schematic (not to exact scale) is:

text
         -- (σ*) A1  <-- antibonding combination of d_z^2 & p_z
         -- (π*) E   <-- antibonding of d_xz/d_yz & p_x/p_y
   NB -->    B1      <-- d_x^2 - y^2
   NB -->    B2      <-- d_xy
         -- (π) E    <-- bonding of d_xz/d_yz & p_x/p_y
         -- (σ) A1   <-- bonding of d_z^2 & p_z

Labels “$\sigma$, \pi), or “NB” (non‐bonding) and the corresponding Mulliken symbols $A_1, E, B_1, B_2$ in $C_{4v}$ round out the diagram.

(iii) Mn–N Bond Order

From the simple MO picture above, a terminal metal–nitrido is typically described as having:

• One strong $\sigma$-bond (M $d_{z^2}$ with N $p_z$),
• Two $\pi$-bonds (M $(d_{xz}, d_{yz})$ with N $(p_x, p_y)$).

Hence the metal–nitrido linkage is conventionally considered to be a triple bond; so the formal bond order is 3.

(iv) Prediction of the EPR Spectrum of $\{PW_{11}O_{39}(Mn-N_3)\}^{5-}$

Here we focus on the azide complex, i.e. before photolysis to the nitrido. From part (i), the manganese in the azide complex is Mn(III), $d^4$. The question is the spin state and the resulting EPR behavior:

1. Likely High‐spin $d^4$.

   • In an octahedral or pseudo‐octahedral environment, high‐spin $d^4$ implies $S=2$.
   • Integer‐spin ($S=2$) complexes often exhibit significant zero‐field splitting (ZFS).

2. Zero‐field splitting and EPR

   • Because $S=2$ is an integer spin, it does not have simple Kramers degeneracy like half‐integer systems (e.g., $S=5/2$ in Mn(II)).
   • Large zero‐field splitting often pushes the EPR transitions outside the usual X‐band range or makes them very broad at room temperature.
   • As a result, many high‐spin Mn(III) complexes are “EPR‐silent” (or show only very broad, weak signals) under standard conditions.
   • If one cools the sample sufficiently and/or uses lower/higher microwave frequencies (Q‐band, etc.), sometimes partial signals are observed—but in a typical undergraduate‐level discussion, the simpler statement is that “the high‐spin $d^4$ Mn(III) complex is usually EPR‐silent at X‐band due to large zero‐field splitting.”

Hence the conventional textbook prediction is that $\{PW_{11}O_{39}(Mn-N_3)\}^{5-}$ (with Mn(III), $d^4$, $S=2$):

• Will be paramagnetic,
• But is likely EPR‐silent (or at most shows a very broad or complex signal) in normal X‐band due to the combination of integer spin and large zero‐field splitting.

Summary of Answers

1. Oxidation state and $d$-count (for the nitrido complex):

   $$\text{Mn is }+5,\quad d^2.$$

2. MO Diagram (tetragonal M≡N in $C_{4v}$):

   • $\sigma$-bond = $d_{z^2}(A_1) + p_z(A_1)$.
   • $\pi$-bonds = $(d_{xz}, d_{yz})(E) + (p_x, p_y)(E)$.
   • $d_{x^2 - y^2}(B_1)$ and $d_{xy}(B_2)$ = nonbonding (no match in symmetry with N p‐orbitals).

3. Mn–N bond order:

   $$3 \quad (\text{one }\sigma\text{ plus two }\pi\text{ bonds}).$$

4. EPR spectrum of $\{PW_{11}O_{39}(Mn-N_3)\}^{5-}$:

   • Mn(III), $d^4$, high‐spin ($S=2$).
   • Large zero‐field splitting typically leads to EPR silence (or only a broad signal) at standard X‐band frequencies.