Medium
For a given atomic orbital,$n = 3$. What are the possible values of $l$ and $m_l$?
(N/A) For $n = 3$,the possible values of the azimuthal quantum number $l$ range from $0$ to $(n - 1)$.
Thus,$l = 0, 1, 2$.
For each value of $l$,the magnetic quantum number $m_l$ ranges from $-l$ to $+l$ including $0$.
If $l = 0$,$m_l = 0$.
If $l = 1$,$m_l = -1, 0, +1$.
If $l = 2$,$m_l = -2, -1, 0, +1, +2$.
Total number of $m_l$ values $= 1 + 3 + 5 = 9$.
Thus,$l = 0, 1, 2$.
For each value of $l$,the magnetic quantum number $m_l$ ranges from $-l$ to $+l$ including $0$.
If $l = 0$,$m_l = 0$.
If $l = 1$,$m_l = -1, 0, +1$.
If $l = 2$,$m_l = -2, -1, 0, +1, +2$.
Total number of $m_l$ values $= 1 + 3 + 5 = 9$.
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The number of correct statements from the following:
$A.$ For $1s$ orbital,the probability density is maximum at the nucleus.
$B.$ For $2s$ orbital,the probability density first increases to maximum and then decreases sharply to zero.
$C.$ Boundary surface diagrams of the orbitals enclose a region of $100\%$ probability of finding the electron.
$D.$ $p$ and $d$-orbitals have $1$ and $2$ angular nodes respectively.
$E.$ Probability density of $p$-orbital is zero at the nucleus.
$A.$ For $1s$ orbital,the probability density is maximum at the nucleus.
$B.$ For $2s$ orbital,the probability density first increases to maximum and then decreases sharply to zero.
$C.$ Boundary surface diagrams of the orbitals enclose a region of $100\%$ probability of finding the electron.
$D.$ $p$ and $d$-orbitals have $1$ and $2$ angular nodes respectively.
$E.$ Probability density of $p$-orbital is zero at the nucleus.
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