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Standard XII

Chemistry

Quantum Numbers

Which of the ...

Question

Which of the following is an impossible set of quantum numbers?

\begin{matrix} n & l & m & s 3 & 2 & 0 & + \frac{1}{2} \end{matrix}

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\begin{matrix} n & l & m & s 3 & 2 & - 2 & + \frac{1}{2} \end{matrix}

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\begin{matrix} n & l & m & s 3 & 2 & - 3 & + \frac{1}{2} \end{matrix}

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\begin{matrix} n & l & m & s 3 & 2 & - 1 & + \frac{1}{2} \end{matrix}

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Solution

The correct option is C $\begin{matrix} n & l & m & s 3 & 2 & - 3 & + \frac{1}{2} \end{matrix}$

Value of $m$ depends on the value of $l$ .
$m$ can be $- l$ to $+ l$ .
So, for $l$ = 2, $m$ can never be -3.

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Similar questions

Q. Consider the value of all four quantum numbers for the last electron and spin multiplicity

(2 s + 1)

,
for the two given elements

^{'} X^{'}

and

^{'} Y^{'}

in their ground state:

\begin{matrix} n & l & m & s & [2 s + 1] X : & 2 & 0 & 0 & + \frac{1}{2} & 1 Y : & 2 & 1 & - 1 & - \frac{1}{2} & 4 \end{matrix}

Then according to given information the correct statement is:

Q. Find modal item of the followings set of numbers:

Q. which of the following set of quantum numbers is not possible
a. n=3, l=1, m=+2
b. n=3, l=1, m=+1
c. n=3, l=0, m=0
d. n=3, l=2, m=+2

Q. Arrange the electrons represented by the following sets of quantum numbers in decreasing order of energy.
(i)

n = 4, l = 0, m_{l} = 0, m_{s} = + \frac{1}{2}

(ii)

n = 3, l = 1, m_{l} = 1, m_{s} = - \frac{1}{2}

(iii)

n = 3, l = 2, m_{l} = 0, m_{s} = + \frac{1}{2}

(iv)

n = 3, l = 0, m_{l} = 0, m_{s} = - \frac{1}{2}

The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination(s) has/have the same energy lists:

1. n = 4, l = 2, m_l = –2 , m_s = –1/2

2. n = 3, l = 2, m_l= 1 , m_s = +1/2

3. n = 4, l = 1, m_l = 0 , m_s = +1/2

4. n = 3, l = 2, m_l = –2 , m_s = –1/2

5. n = 3, l = 1, m_l = –1 , m_s= +1/2

6. n = 4, l = 1, m_l = 0 , m_s = +1/2

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Which of the following is an impossible set of quantum numbers?

The correct option is C nlms32−3+12 Value of m depends on the value of l. m can be −l to +l. So, for l= 2, m can never be -3.

The correct option is C $\begin{matrix} n & l & m & s 3 & 2 & - 3 & + \frac{1}{2} \end{matrix}$

Value of $m$ depends on the value of $l$ .
$m$ can be $- l$ to $+ l$ .
So, for $l$ = 2, $m$ can never be -3.