**Hint:**

For an atomic orbital with principal quantum number n, the possible values of azimuthal quantum number l are 0, 1, 2, ..., (n-1).

For each value of l, the possible values of magnetic quantum number ml are -l, -l+1, ..., 0, ..., l-1, l.

3d orbital has 5 possible ml values: -2, -1, 0, 1, 2.

1p, 2p, and 2s are possible orbitals, but 3f is not.

**Solution:**

**(i) Given that n = 3:**

For a specified principal quantum number $ n $, the azimuthal quantum number $l $ can range from 0 up to $ n - 1 $.

So, for $n = 3$:

$l$ can be 0, 1, or 2.

Each value of $l$ allows for different magnetic quantum numbers $m_l$:

When $l = 0, \quad m_l = 0$.

When $l = 1, \quad m_l$ can be $-1, 0$, or $1$.

When $l = 2, \quad m_l$ can range from $-2$ to $2$ (i.e., -2, -1, 0, 1, 2).

**(ii) For the 3d orbital:**

The $d$ orbital corresponds to $l = 2$. For each $l$ value, there are $2l + 1$ possible values for $m_l$. Thus:

For $l = 2$, $m_l$ can be -2, -1, 0, 1, or 2.

**(iii) Evaluation of possible orbitals:**

1p: This doesn't exist because the minimum principal quantum number $n$ for a $p$ orbital (where $l = 1$) is 2.

2s: Exists and is valid.

2p: Exists and is valid.

3f: This doesn't exist. An $f$ orbital corresponds to $l = 3$, but for $l = 3$, the minimum value of $n$ would be 4. Hence, the 3f configuration isn't possible.

**Concisely:**

The 3d, 2s, and 2p orbitals are valid configurations.

The 1p and 3f orbitals are not valid configurations for the given quantum numbers.

**Note: **

Quantum Numbers and Orbital Properties

For n=3 orbital, possible l values: 0, 1, 2

For each l, possible ml values: -l to +l

3d orbital: ml = -2, -1, 0, 1, 2

Possible orbitals: 1p, 2s, 2p; 3f not possible