1. Identify the property asked

We need the metal among $\mathrm{Cr}, \mathrm{Co}, \mathrm{Fe}$ and $\mathrm{Ni}$ which has the lowest enthalpy of atomisation, and then find its number of valence electrons.

2. Use the trend for enthalpy of atomisation in 3d-series

For transition metals, enthalpy of atomisation generally depends on the strength of metallic bonding, which is related to the number of unpaired electrons and participation of both $3d$ and $4s$ electrons.

Among the given elements:

• $\mathrm{Cr}$ : relatively lower atomisation enthalpy than $\mathrm{Fe}, \mathrm{Co}, \mathrm{Ni}$
• $\mathrm{Fe}, \mathrm{Co}, \mathrm{Ni}$ : have stronger metallic bonding and hence higher atomisation enthalpy

Thus, the metal with the lowest enthalpy of atomisation among these is:

$\boxed{\mathrm{Cr}}$

3. Find the valence electrons of chromium

Electronic configuration of chromium is:

$\mathrm{Cr} : [Ar] \, 3d^5 4s^1$

For transition elements, valence electrons are taken as:

$ns + (n-1)d$

So for chromium:

$3d^5 + 4s^1 = 6 \text{ valence electrons}$

Hence, the required number is:

$\boxed{6}$

4. Match with the options

• A: $10$
• B: $6$
• C: $9$
• D: $8$

Therefore, the correct option is:

$\boxed{\text{B}}$

5. Compare with stored correct answer

Stored correct answer is B.

Our derived answer is also B.