1D.21
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1D.21
Write the subshell notation (3d, for instance) and the number of orbitals having the following quantum numbers: (a) n 5 5, l52; (b) n51,l50;(c) n56,l53; (d) n52,l51.
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Re: 1D.21
i think the question was mistyped but just for reference:
Write the subshell notation (3d, for instance) and the number of orbitals having the following quantum numbers:
a. n=5, l=2 (5d, five)
b. n=1, l=0 (1s, one)
c. n=6, l=2 (6f, seven)
d. n=7, l=3 (2p, three)
Know that
l = 0 corresponds to the s-orbital
l = 1 corresponds to the p-orbital
l = 2 corresponds to the d-orbital
l = 3 corresponds to the f-orbital
Know that
s-orbital = 1 orbital, total of 2 e-
p-orbital = 3 orbitals, total of 6e-
d-orbital = 5 orbitals, total of 10e-
f-orbital = 7 orbitals, total of 14e-
Each orbital has 2 e-
Example using a (n=5, l=2)
--> Since l=2, we can assume this is a d-orbital. We know that the d-orbital has 5 orbitals, each containing 2 electrons
Write the subshell notation (3d, for instance) and the number of orbitals having the following quantum numbers:
a. n=5, l=2 (5d, five)
b. n=1, l=0 (1s, one)
c. n=6, l=2 (6f, seven)
d. n=7, l=3 (2p, three)
Know that
l = 0 corresponds to the s-orbital
l = 1 corresponds to the p-orbital
l = 2 corresponds to the d-orbital
l = 3 corresponds to the f-orbital
Know that
s-orbital = 1 orbital, total of 2 e-
p-orbital = 3 orbitals, total of 6e-
d-orbital = 5 orbitals, total of 10e-
f-orbital = 7 orbitals, total of 14e-
Each orbital has 2 e-
Example using a (n=5, l=2)
--> Since l=2, we can assume this is a d-orbital. We know that the d-orbital has 5 orbitals, each containing 2 electrons
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- Posts: 100
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Re: 1D.21
For reference for l,
l=0 ---> s orbital
l=1 ---> p orbital
l=2 ---> d orbital
l=3 ---> f orbital (we don't really need to know this)
the principal quantum number, n, just needs to be in the front so,
n=5 l=2 ---> 5d
n=1 l=0 ---> 1s
n=6 l=3 ---> 6f
n=2 l=1 ---> 2p
l=0 ---> s orbital
l=1 ---> p orbital
l=2 ---> d orbital
l=3 ---> f orbital (we don't really need to know this)
the principal quantum number, n, just needs to be in the front so,
n=5 l=2 ---> 5d
n=1 l=0 ---> 1s
n=6 l=3 ---> 6f
n=2 l=1 ---> 2p
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