The secondary valency and the number of hydrogen bonded water molecule(s) in CuSO₄·5H₂O, respectively, are:

🧠 The Werner Picture of Blue Vitriol

$\mathrm{CuSO_4 \cdot 5H_2O}$ is structurally $[\mathrm{Cu(H_2O)_4}]\mathrm{SO_4 \cdot H_2O}$:

• 4 waters sit in the equatorial plane of $\mathrm{Cu^{2+}}$ (square-planar coordination).
• 1 water sits outside the coordination sphere, hydrogen-bonded to two coordinated waters and to the sulfate.
• (Sulfate provides 2 weak axial Cu–O contacts in the crystal, but these are not counted in Werner's "secondary valency" by NCERT convention.)

So:

• Secondary valency (Werner's term for primary coordination) = 4.
• H-bonded waters = 1 (the lattice water).

🗺️ The 4+1 Anchor

Memorise the picture: 4 waters bound to Cu, 1 water in the lattice. The "5" in $\mathrm{CuSO_4 \cdot 5H_2O}$ splits as $4 + 1$.

⚡ Why Square-Planar?

$\mathrm{Cu^{2+}}$ is $\mathrm{d^9}$. Strong Jahn–Teller distortion elongates two trans bonds so much that the equatorial plane dominates → effectively square-planar with weak axial contacts. The four waters in the plane give CN = 4 by Werner's tally.

⚠️ Don't Add Up to 5

The secondary valency is not 5. The 5th water is hydrogen-bonded only — it's not a primary Cu–O bond. Similarly, secondary valency is not 6 even though the actual CN with axial sulfate contacts is sometimes given as 6 in advanced texts; for JEE, use the NCERT picture: 4.

$\boxed{\text{Answer: (2) 4 and 1}}$