The correct order of increasing energy of ato | vedclass

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(B) According to the $(n+l)$ rule,the energy of an orbital increases with the value of $(n+l)$.For $5p$: $n=5, l=1$,so $(n+l) = 6$.For $6s$: $n=6, l=0

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$5p < 6s < 4f < 5d$

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(B) According to the $(n+l)$ rule,the energy of an orbital increases with the value of $(n+l)$.For $5p$: $n=5, l=1$,so $(n+l) = 6$.For $6s$: $n=6, l=0$,so $(n+l) = 6$.For $4f$: $n=4, l=3$,so $(n+l) = 7$.For $5d$: $n=5, l=2$,so $(n+l) = 7$.If $(n+l)$ values are equal,the orbital with the lower $n$ value has lower energy.Comparing $5p$ and $6s$: both have $(n+l) = 6$,but $5p$ has a lower $n$ value,so $5p Comparing $4f$ and $5d$: both have $(n+l) = 7$,but $4f$ has a lower $n$ value,so $4f Thus,the correct order is $5p < 6s < 4f < 5d$.

Column-$I$	Column-$II$
$(a)$ $4s$	$(p)$ Circular orbit around nucleus
$(b)$ $4p$	$(q)$ Non-directional orbital
$(c)$ $1s$	$(r)$ Angular momentum $= 2h/\pi$
$(d)$ $3d$	$(s)$ Radial node is zero

The correct order of increasing energy of atomic orbitals is

Similar Questions

Match the columns and choose the correct option:

Column-$I$ Column-$II$

$(a)$ $4s$ $(p)$ Circular orbit around nucleus

$(b)$ $4p$ $(q)$ Non-directional orbital

$(c)$ $1s$ $(r)$ Angular momentum $= 2h/\pi$

$(d)$ $3d$ $(s)$ Radial node is zero

The atom of the element having atomic number $14$ should have:

The $71^{st}$ electron of an element $X$ with an atomic number of $71$ enters into the orbital:

The number of electrons that can be accommodated in $d_{z^2}$ orbital is

What is the maximum number of electrons which can be accommodated in an atom in which the highest principal quantum number value is $4$?

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