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(D) For any orbital,the principal quantum number $n$ must be greater than the azimuthal quantum number $l$. Since $l = 4$ for $g$ orbitals,the lowest

Vedclass · Accepted Answer

$n = 5, m_l = -4$ to $+4$,no. of electrons $= 18$

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(D) For any orbital,the principal quantum number $n$ must be greater than the azimuthal quantum number $l$. Since $l = 4$ for $g$ orbitals,the lowest possible value for $n$ is $n = l + 1 = 4 + 1 = 5$. The magnetic quantum number $m_l$ ranges from $-l$ to $+l$. For $l = 4$,$m_l$ can take values $\{-4, -3, -2, -1, 0, +1, +2, +3, +4\}$. The number of orbitals in a subshell is given by $2l + 1$. For $l = 4$,there are $2(4) + 1 = 9$ orbitals. Each orbital can hold a maximum of $2$ electrons,so the total number of electrons is $9 	imes 2 = 18$.

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