= Solution
Addition of two spin-one-half angular momenta gives total <spin> $S=0$ or $S=1$. The normalized <singlet state> and <triplet states> are
$$
|0,0\rangle=\frac{|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle}{\sqrt2},
\qquad
|1,1\rangle=|\uparrow\uparrow\rangle,\quad
|1,0\rangle=\frac{|\uparrow\downarrow\rangle+|\downarrow\uparrow\rangle}{\sqrt2},
\quad |1,-1\rangle=|\downarrow\downarrow\rangle.
$$
They have $S^2$ <eigenvalues> $0$ and $2\hbar^2$, and $S_z=M\hbar$. The <singlet state> is antisymmetric under a <particle exchange operator>, whereas all three triplets have symmetric spin-exchange parity. For two identical <electrons> the full state must be antisymmetric: a <singlet state> pairs with a spatial state with symmetric exchange parity, a <triplet state> with an antisymmetric spatial state. Thus \b[two <electrons> in the same <atomic orbital> must form a <singlet state>].
The identity $\mathbf S_1\cdot\mathbf S_2=(S^2-S_1^2-S_2^2)/2$ gives $-3\hbar^2/4$ in the <singlet state> and $+\hbar^2/4$ in the <triplet state>. Hence an <exchange interaction> distinguishes their <energies>, with ordering determined by its coupling sign. The <singlet state> has perfect opposite-spin correlations along any common measurement axis and is <entangled>. The $M=0$ <triplet state> is also <entangled>, while the $M=\pm1$ <triplet states> are <product states> along the chosen axis.
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