For \( n = 4 \), the possible subshells are :
4s (1 orbital)
4p (3 orbitals)
4d (5 orbitals)
4f (7 orbitals)
Among these, only one orbital from each subshell can have a magnetic quantum number \( m_l = 0 \). Therefore, there is one such orbital from 4s, one from 4p, one from 4d, and one from 4f.
Thus, the total number of orbitals with \( n = 4 \) and \( m_l = 0 \) is 4.
Which of the following statement is true with respect to H\(_2\)O, NH\(_3\) and CH\(_4\)?
(A) The central atoms of all the molecules are sp\(^3\) hybridized.
(B) The H–O–H, H–N–H and H–C–H angles in the above molecules are 104.5°, 107.5° and 109.5° respectively.
(C) The increasing order of dipole moment is CH\(_4\)<NH\(_3\)<H\(_2\)O.
(D) Both H\(_2\)O and NH\(_3\) are Lewis acids and CH\(_4\) is a Lewis base.
(E) A solution of NH\(_3\) in H\(_2\)O is basic. In this solution NH\(_3\) and H\(_2\)O act as Lowry-Bronsted acid and base respectively.
Regarding the molecular orbital (MO) energy levels for homonuclear diatomic molecules, the INCORRECT statement(s) is (are):
Let $ f(x) = \begin{cases} (1+ax)^{1/x} & , x<0 \\1+b & , x = 0 \\\frac{(x+4)^{1/2} - 2}{(x+c)^{1/3} - 2} & , x>0 \end{cases} $ be continuous at x = 0. Then $ e^a bc $ is equal to
Total number of nucleophiles from the following is: \(\text{NH}_3, PhSH, (H_3C_2S)_2, H_2C = CH_2, OH−, H_3O+, (CH_3)_2CO, NCH_3\)