a) d^{7} means the configuration. In other words, there will be 3 unpaired electrons of a **3/2 spin**.

b) The correct answer is s- orbital

- When l = 0, it is evident from that the angular momentum of the electron is zero.
- The atomic orbitals which describe these states of zero angular momentum are called s orbitals.
- The s orbitals are distinguished from one another by stating the value of n, the principal quantum number.

c) The correct option is **c**

After the 4s is full we put the remaining six electrons in the 3d orbital and end with 3d9. Therefore the expected electron configuration for Copper will be 1s^{2} 2s^{2} 2p^{2} 3s^{2} 3p^{2} 4s^{1} 3d^{10}.

**[Ar] 4s**^{1} 3d^{10}