Question

the first element of the group in many ways differs from the others heavier members of the group. This is due to :
A. The small size
B. The high electronegativity and high ionisation potential
C. The unavailability of d-orbitals
D. All of the above

Answer 1

Hint: The first element of each group in s and p-block anticipating noble gases differentiate in many regards from the other members of their respective group.This property is called as ANOMALOUS Property.

Complete step by step answer:
- Option 1st: Since the nuclear radii decline over a period, the p-block molecules are more modest than their closest s or d block particles; consequently F molecule has the littlest span. Related with little molecules the \[2p\] orbitals are minimal and impact the bonds shaped. Interelectronic shocks are in this way more critical in \[2p\] than in np orbitals (where \[n>2\]). These outcomes in the \[N-N\], \[O-O\] and \[F-F\] bonds are nearly more vulnerable than the \[P-P\] and \[Cl-Cl\] bonds separately.

- Option 2nd: Electronegativity is characterized as a proportion of the capacity of a particle to draw in the mutual electron pair in a covalent cling to itself. Electronegativity increments along the period and diminishes down the gathering. Fluorine is the most electronegative of the apparent multitude of components. The second generally electronegative component is oxygen trailed by nitrogen in the third position.

- Option 3rd: Absence of d-orbital: The absence of d-orbitals in the elements of 2nd period and the presence of d orbital in the heavier elements.Using these d-orbitals, the elements of 3rd period can accommodate more electron and hence can expand their covalency beyond 4. Since d-orbitals are higher.

The first member of each group differs from the heavier elements in its ability to form pn-pn multiple bonds either with itself or with the other elements of the second period. This type of n-bonding is not strong in case of heavier p-block elements. The heavier elements also form n-bonds but this involves d-orbitals.
A) The smaller size of the first element .
B) The higher electronegativity and higher ionization potential of the first element.
C) The unavailability of d-orbitals in the first element.
D) But not because of the higher shielding effect.
 Hence option A,B & C are correct.

Therefore the correct option is D. All of the above

Note:
Thus, it has obviously been set up that the second-time span components are extraordinary. Indeed, they show intermittent properties that are like the second component of the following gathering (for example Lithium is like Magnesium and Beryllium to Aluminum) or at the end of the day, they have an askew relationship.