Group 15 elements (such as nitrogen, phosphorus, arsenic, etc.) can form covalent bonds with other elements, and the maximum covalency typically corresponds to the number of bonds that can be formed.
For a non-metallic element in group 15, the maximum covalency is usually determined by the number of available orbitals for bonding.
- Nitrogen, the lightest element in Group 15, has the weakest E-E bond due to its small atomic size and high electronegativity, which limits the number of bonds it can form.
- Phosphorus, arsenic, and other heavier elements in the group can form a maximum of 4 bonds due to their larger atomic size and availability of d-orbitals for bonding.
Thus, the correct answer is that the maximum covalency for a non-metallic Group 15 element with the weakest E-E bond is 4.
Column I | Column II |
P. Solubility | i. Reverse phase chromatography |
Q. Ionic charge | ii. Ultracentrifugation |
R. Polarity | iii. Salting out |
S. Molecular size | iv. Isoelectric focusing |
v. Gel electrophoresis |
Column I | Column II |
P. Solubility | i. Reverse phase chromatography |
Q. Ionic charge | ii. Ultracentrifugation |
R. Polarity | iii. Salting out |
S. Molecular size | iv. Isoelectric focusing |
v. Gel electrophoresis |
Let \( A = \{-3, -2, -1, 0, 1, 2, 3\} \). A relation \( R \) is defined such that \( xRy \) if \( y = \max(x, 1) \). The number of elements required to make it reflexive is \( l \), the number of elements required to make it symmetric is \( m \), and the number of elements in the relation \( R \) is \( n \). Then the value of \( l + m + n \) is equal to:
For hydrogen-like species, which of the following graphs provides the most appropriate representation of \( E \) vs \( Z \) plot for a constant \( n \)?
[E : Energy of the stationary state, Z : atomic number, n = principal quantum number]
The number of 6-letter words, with or without meaning, that can be formed using the letters of the word MATHS such that any letter that appears in the word must appear at least twice, is $ 4 \_\_\_\_\_$.