Nickel atom can lose two electrons to form $\mathrm{Ni}^{2+}$ ion. The atomic number of nickel is 28 . From which orbital will nickel lose two electrons?
${ }_{28} \mathrm{Ni}=1 s^2, 2 s^2, 2 p^6, 3 s^2, 3 p^6, 3 d^8, 4 s^2$; Nickel will lose 2 electrons from $4 s$ (outermost shell) to form $\mathrm{Ni}^{2+}$ ion.
Hence, $${ }_{28} \mathrm{Ni}^{2+}=1 s^2, 2 s^2, 2 p^6, 3 s^2, 3 p^6, 3 d^8, 4 s^0$$
$$\begin{aligned} &\text { Which of the following orbitals are degenerate? }\\ &3 d_{x y}, 4 d_{x y}, 3 d_{z^2}, 3 d_{y z}, 4 d_{y z}, 4 d_{z^2} \end{aligned}$$
The orbitals which belongs to same subshell and same shell are called degenerate orbitals. $\left(3 d_{x y}, 3 d_{z^2}, 3 d_{y z}\right)$ and $\left(4 d_{x y}, 4 d_{y z}, 4 d_{z^2}\right)$ are the two sets of degenerate orbitals.
Calculate the total number of angular nodes and radical nodes present in $3 p$ orbital.
For $3 p$-orbital, principal quantum number, $n=3$ and azimuthal quantum number $l=1$
Number of angular nodes $=l=1$
Number of radial nodes $=n-l-1=3-1-1=1$
The arrangement of orbitals on the basis of energy is based upon their $(n+l)$ value. Lower the value of $(n+l)$, lower is the energy. For orbitals having same values of $(n+l)$, the orbital with lower value of $n$ will have lower energy.
I. Based upon the above information, arrange the following orbitals in the increasing order of energy.
(a) $1 s, 2 s, 3 s, 2 p$
(b) $4 s, 3 s, 3 p, 4 d$
(c) $5 p, 4 d, 5 d, 4 f, 6 s$
(d) $5 f, 6 d, 7 s, 7 p$
II. Based upon the above information. Solve the questions given below.
(a) Which of the following orbitals has the lowest energy?
$$4 d, 4 f, 5 s, 5 p$$
(b) Which of the following orbitals has the highest energy?
$$5 p, 5 d, 5 f, 6 s, 6 p$$
I. (a) $(n+l)$ values of $1 s=1+0=1,2 s=2+0=2,3 s=3+0=3,2 p=2+1=3$
Hence, increasing order of their energy is
$$1 s<2 s<2 p<3 s$$
(b) $4 s=4+0=4,3 s=3+0=3,3 p=3+1=4, 4 d=4+2=6$
Hence, $3 s<3 p<4 s<4 d$
(c) $5 p=5+1=6,4 d=4+2=6,5 d=5+2=7,4 f=4+3=7,6 s=6+0=6$
Hence, $4 d<5 p<6 s<4 f<5 d$
(d) $5 f=5+3=8,6 d=6+2=8,7 s=7+0=7,7 p=7+1=8$.
Hence, $$7 s<5 f<6 d<7 p$$
II. (a) $(n+l)$ values of $4 d=4+2=6,4 f=4+3=7,5 s=5+0=5,7 p=7+1=8$
Hence, $5 s$ has the lowest energy.
$$\text { (b) } 5 p=5+1=6,5 d=5+2=7,5 f=5+3=8,6 s=6+0=6,6 p=6+1=7$$
Hence, $5f$ has highest energy.
Which of the following will not show deflection from the path on passing through an electric field? Proton, cathode rays, electron, neutron.
Neutron being neutral will not show deflection from the path on passing through an electric field.
Proton, cathode rays and electron being the charged particle will show deflection from the path on passing through an electric field.