The out-of-plane displacements in the wag transform as $y$, hence as $B_2$. A one-quantum combination of two distinct normal modes transforms as the product of their irreducible representations. The observed feature satisfiesA combination of the wag with an $A_1$ stretch has symmetry $B_2\otimes A_1=B_2$ and can absorb through ${\mu }_y$. In contrast, $B_2\otimes B_1=A_2$ has no electric-dipole component in $C_{2v}$ and is forbidden in the stated symmetry model. The missing feature near $4010=1167+2843\mathrm{c}{\mathrm{m}}^{-1}$ therefore identifies the $2843$ stretch as $B_1$.

The completed assignments areThe allowed combination obtains intensity through anharmonic/dipole-coupling terms; a purely separable harmonic model with a strictly linear dipole has no combination bands. The argument uses both the observed combination and the selection rule: absence of a feature by itself would not generally prove a forbidden symmetry.^[ref]

Let $Q_k$ be a vibrational normal mode coordinate with symmetry ${\Gamma }_k$. The harmonic vibrational ground state is totally symmetric; the one-quantum state transforms as $Q_k$. Integration over a symmetry-invariant domain annihilates any integrand lacking the totally symmetric representation. Thus a fundamental requiresfor some Cartesian component $\alpha$. For the one-dimensional irreducible representations used here, this means ${\Gamma }_k=\Gamma (x)$, $\Gamma (y)$ or $\Gamma (z)$. Equivalently, expanding the electric dipole moment givesthe derivative can be nonzero only if $Q_k$ and ${\mu }_{\alpha }$ have compatible symmetry. This is the symmetry criterion for vibrational infrared activity; it permits a transition without guaranteeing a large intensity.

A useful vibration-rotation expression, with energies measured from the potential minimum, isHere $v=0,1,\dots$ is the vibrational quantum number, $J=0,1,\dots$ the rotational angular-momentum quantum number, ${\tilde{\nu }}_e$ the harmonic spectroscopic wavenumber, $x_e$ a dimensionless anharmonic coefficient, and $B_v$ the vibrational-state-dependent rotational constant. $B_e$ is its equilibrium value and ${\alpha }_e$ describes vibration-rotation coupling. All wavenumber constants use the same inverse-length units; $h$ is the Planck constant and $c$ the speed of light.

The elementary harmonic rigid rotor result follows by setting $x_e={\alpha }_e=0$, with ${\tilde{\nu }}_e=(2\pi c{)}^{-1}\sqrt{k/\mu }$ and $B_e=h/(8{\pi }^{2}cI)$, where $I=\mu r_e^2$. The molecule's electronic energy would add a common constant.

The closed argon core contributes $L=S=0$. In the half-filled $3d^5$ shell, Hund's rules put one parallel-spin electron in each $m_{\ell }=-2,-1,0,1,2$ orbital. This gives $S_d=5/2$ and the orbital term $L_d=0$. Adding the parallel $4s$ electron, with $\ell =0$ and $s=1/2$, givesThe spin multiplicity is seven. The magnetic projection labels run $M_S,M_J=-3,-2,\dots ,3$, while $M_L=0$. The term symbol agrees with the observed neutral-chromium ground level.^[ref]

The $3d$ and $4s$ atomic orbitals of chromium are close enough in energy that electron-electron terms can overturn the simple filling order. The configuration $[\mathrm{Ar}]3d^{5}4s^1$ permits five singly occupied $3d$ orbitals and a parallel-spin $4s$ electron. Relative to $3d^{4}4s^2$, it removes the $4s$ pair and permits more parallel-spin exchange interactions, lowering the Coulomb-and-exchange contribution enough to offset the orbital promotion cost.

As qualitative bookkeeping, six parallel electron spins give $\left(\genfrac{}{}{0ex}{}{6}{2}\right)=15$ same-spin pairs, compared with $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$ in high-spin $3d^{4}4s^2$. The individual exchange integrals are unequal, so this is an explanation of the tendency, rather than a quantitative atomic energy calculation. The observed ground state configuration is $\boxed{[\mathrm{Ar}]3d^{5}4s^1}$.^[ref]

If two fermions occupied the same complete spin orbital $u$, their antisymmetrized product would beThe zero vector cannot represent a normalized quantum state. This is the Pauli exclusion principle: at most one identical fermion occupies any complete one-particle state. Two electrons may occupy the same spatial orbital with opposite spins, because $a\alpha$ and $a\beta$ are different spin orbitals.