Question

Which of the following is a correct set of quantum numbers for an electron in a 3d orbital? on = 3,1 = 0, mi = -1, ms = +1/2 on = 3,1 = 2, mj = +3, ms = +1/2 ОО n = 3, 1 = 1, mi = +3, ms = +1/2 on=3,1 = 3, mi = +2, ms = +1/2 on = 3, I = 2, mi = -2, ms = +1/2

Answer 1

The principal quantum number, signified by n, is the main energy level occupied by the electron.  They are described in whole number increments (e.g., 1, 2, 3, 4, 5, 6, ...).

The angular momentum quantum number, signified by l, describes the orbital shape. The angular momentum quantum number can have positive values of zero to (n−1). For instance, if n=2, l could be either 0 or 1.

The magnetic quantum number, signified as m, describes the orbital orientation in space. For a given value of the angular momentum quantum number, l, there can be (2l+1) values for m.

The spin quantum number, signified as m_s, describes the spin for a given electron. An electron can have one of two associated spins, (+1/2) spin, or (−1/2) spin. An electron cannot have zero spin.

For 3d subshell, the value of n is 3.

So, l can have values 0, 1, and 2. For d subshell, value of l = 2

Similarly, m can have values -2, -1, 0, 1, and 2. If the value of l=2, the possible values of m are +2 and -2.

And m_s can have only two values +1/2 and -1/2.

Considering the four points above, the only option that matches the requirements is Option 5 (n=3, l=2, m=-2, s=+1/2)

Other options:

Option 1 is incorrect because the value of l for d subshell is 2, not 0.

Option 2 is incorrect because the value of m cannot be +3, if the value of l is 2.

Option 3 is incorrect because the value of l for d subshell is 2, not 1

Option 1 is incorrect because value of l for d subshell is 2, not 3.

Answer : Option 5

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