Explanation:

Ti has an atomic number of 22, and its ground state electronic configuration is [Ar] 4s²3d². The excited state configuration given is [Ar] 4s²3d¹4p¹.

Step 1: Determine the number of electrons with spin up and spin down.

In the excited state configuration [Ar] 4s²3d¹4p¹, there are 2 electrons in the 4s subshell with opposite spins, 5 electrons in the 3d subshell with opposite spins, and 1 electron in the 4p subshell with spin up.

Step 2: Determine the number of microstates with zero total spin.

For a system with N electrons, the total spin S can range from -S to +S in integer steps. The number of microstates with zero total spin is given by:

(number of microstates with S=0) = [(2S + 1) / (N + 1)] * (N choose N/2)

where N choose N/2 is the binomial coefficient, which gives the number of ways to choose N/2 electrons to have spin up.

For the excited state configuration [Ar] 4s²3d¹4p¹, the number of microstates with zero total spin is:

(number of microstates with S=0) = [(2*0 + 1) / (8)] * (8 choose 4) = 70 / 70 = 1

Step 3: Determine the number of microstates with non-zero total spin.

The number of microstates with non-zero total spin is given by:

(number of microstates with S>0) = (2S + 1) * [(N choose (N/2) + 1) / (N + 1)]

For the excited state configuration [Ar] 4s²3d¹4p¹, the number of microstates with non-zero total spin is:

(number of microstates with S>0) = (2*1 + 1) * [(8 choose 5) / 9] = 15

Step 4: Add the number of microstates with zero and non-zero total spin.

The total number of microstates for the excited state configuration [Ar] 4s²3d¹4p¹ is:

(number of microstates) = (number of microstates with S=0) + (number of microstates with S>0) = 1 + 15 = 16

Therefore, the answer is (b) 15.