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(A) In a hydrogen atom,the energy of an orbital depends only on the principal quantum number $(n)$.For a given $n$,all orbitals are degenerate (have t

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$B, C$ and $D$ only

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(A) In a hydrogen atom,the energy of an orbital depends only on the principal quantum number $(n)$.For a given $n$,all orbitals are degenerate (have the same energy).The given orbitals have the following $n$ values:$A$. $4s$ $(n=4)$$B$. $3p_x$ $(n=3)$$C$. $3d_{x^2-y^2}$ $(n=3)$$D$. $3d_{z^2}$ $(n=3)$$E$. $4p_z$ $(n=4)$Since the energy increases with $n$,the orbitals with $n=3$ $(3p_x, 3d_{x^2-y^2}, 3d_{z^2})$ have the lowest energy.Therefore,the correct option is $B, C$ and $D$ only.

For a hydrogen atom,the orbital$(s)$ with the lowest energy is/are:
$A$. $4s$
$B$. $3p_x$
$C$. $3d_{x^2-y^2}$
$D$. $3d_{z^2}$
$E$. $4p_z$
Choose the correct answer from the options given below:

Similar Questions

$A$: The energy of an electron can be determined by the principal quantum number.
$R$: The principal quantum number $(n)$ determines the possible distance of the electron from the nucleus.

The configuration $1s^2 2s^2 2p^5 3s^1$ shows:

The correct decreasing order of energy for the orbitals having the following set of quantum numbers:
$A$. $n=3, \ell=0, m=0$
$B$. $n=4, \ell=0, m=0$
$C$. $n=3, \ell=1, m=0$
$D$. $n=3, \ell=2, m=1$

The principal quantum number represents:

Which of the following sets of quantum numbers is not possible?

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For a hydrogen atom,the orbital$(s)$ with the lowest energy is/are:$A$. $4s$$B$. $3p_x$$C$. $3d_{x^2-y^2}$$D$. $3d_{z^2}$$E$. $4p_z$Choose the correct answer from the options given below: