Given:(A) n = 5, m_l = +1(B) n = 2, l = 1, m_ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) $(A)\ n = 5$ implies $l$ can be $0, 1, 2, 3, 4$ ($s, p, d, f, g$ orbitals).For a given $m_l = +1$,the orbitals that can have this value are $p$ $(

Vedclass · Accepted Answer

$8$ and $1$

Vedclass · Answer

(B) $(A)\ n = 5$ implies $l$ can be $0, 1, 2, 3, 4$ ($s, p, d, f, g$ orbitals).For a given $m_l = +1$,the orbitals that can have this value are $p$ $(l=1)$,$d$ $(l=2)$,$f$ $(l=3)$,and $g$ $(l=4)$.Each orbital can hold $2$ electrons $(m_s = +1/2, -1/2)$.Total electrons $= 2 (p) + 2 (d) + 2 (f) + 2 (g) = 8$ electrons.$(B)\ n = 2, l = 1, m_l = -1, m_s = -1/2$ represents a specific electron in the $2p$ orbital.Since all four quantum numbers are specified,it corresponds to exactly $1$ electron.

Given:
$(A)\ n = 5, m_l = +1$
$(B)\ n = 2, l = 1, m_l = -1, m_s = -1/2$
The maximum number of electron$(s)$ in an atom that can have the quantum numbers as given in $(A)$ and $(B)$ are respectively:

Similar Questions

The number of orbitals associated with quantum numbers $n=5, m_s=+\frac{1}{2}$ is

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:

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

The number of orbitals in the fourth principal quantum number $(n = 4)$ is:

Choose the pair whose ions have the similar electronic configuration.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Given:$(A)\ n = 5, m_l = +1$$(B)\ n = 2, l = 1, m_l = -1, m_s = -1/2$The maximum number of electron$(s)$ in an atom that can have the quantum numbers as given in $(A)$ and $(B)$ are respectively: