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i. Write the electronic configurations o...

i. Write the electronic configurations of the following ions:
a. `H^(Θ)`, b. `Na^(o+)`, c. `O^(2-)`, d. `F^(Θ)`
ii. What are the atomic numbers of elements whose outermost electrons are represented by
a. `3s^(1)`, b. `2p^(3)`, c. `3p^(5)`?
iii. Which atoms are indicated by the following configurations?
a. `[He]2s^(1)`, b. `[Ne]3s^(2) 3p^(3)`, c. `[Ar]4s^(2) 3d^(1)`

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` (a) H^(-) ion`
the elecrtrons configuration of H atom is `1s^(1)`
A negative charge on the species indicates the gain of an electrons by it
Electronic configuration of ` H^(-) = 1s^(2)`
` Na^(+) ion `
the electronic congifuration of Na atom is ` 1s^(2) 2s^(2) 2p^(6) 3s^(0)or 1s^(2)2s^(2) 2p^(6)`
(c). `O^(2-) ion`
the electronic confuration of 0 atom is `1s^(2) 2s^(2) 2p^(4)`
A diengative charge on the species indicates that two electrons are gianed by it .
Electronic congiguration of `O^(2-) "ion" = 1s^(2)2s^(2) 2p^(2)`
(d) `F^(-)`ion
the electronic configuration of F atom is ` 1s^(2) 2s^(2) 2p^(6)`
(ii) (a) `3s^(1)`
conmpletiing the electron configuration of the element as
`1s^(2) 2s^(2) 2p^(6)3s^(1)`
number of electrons present in the atom of the element
2+2 + 6+ 1=11
number of electrons present in the atom of the element = 2+ 2+ 3=7
Atomic number of the element = 7
`3p^(5)`
completing the electron configuration of the element as
1s2 2s2 2p6 3s5, number of electrons present in the atom of the element = 2 + 2+ 6 + 2+ 5 = 17
Atomic number of the element = 17
(iii)` [ He] 2s^(1)`
the electronic configuration of the element is ` [ He] 2s^(1) = 1s^(2) 2s^(1)`
atomic number of the element = 3
Hence , the element with the electrons configuration ` [He] 2s^(1) = 1 s^(2) 2s^(1)`
atomic number of the element
Hence ,, the element with the electronic configuration ` [He]2s^(1) = 1s^(2)2s^(1)` is litihium (Li)
`[Ar]4s^(2)3d^(1)`
the electronic configuration of the element is ` [Ar] 4s^(1) 3d^(1) = 1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 4s^(2) 3d^(1)`
Atomic number of the element = 21
hence the element with the electrons configuration `[ Ar] 4s^(1) 3d^(1)` is scandium (Sc).
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