Answers:
(a) 1s² 2s²2p³; (b) 1s² 2s²2p⁶ 3s²3p⁶ 4s²3d²; (c) 1s² 2s²2p⁶ 3s²3p⁵
Step-by-step explanation:
One way to solve this problem is to add electrons to the orbitals one-by-one until you have added the required amount.
Fill the subshells in the order listed in the diagram below. Remember that an s subshell can hold two electrons, while a p subshell can hold six, and a d subshell can hold ten.
(a) <em>Seven electrons
</em>
1s² 2s²2p³
There are two electrons in the 2s subshell and three in the 2p subshell. The remaining two electrons are in the inner 1s subshell.
(b) <em>22 electrons
</em>
1s² 2s²2p⁶ 3s²3p⁶ 4s²3d²
There are two electrons in the 4s subshell and two in the 2p subshell. The remaining 18 electrons are in the inner subshells.
(c) <em>17 electrons</em>
1s² 2s²2p⁶ 3s²3p⁵
There are two electrons in the 3s subshell and five in the 2p subshell. The remaining 10 electrons are in the inner subshells.
Answer:
% of n-propyl chloride = 43.48 %
Explanation:
There are 2 secondary hydrogens and 6 primary hydrogens
The rate of abstraction of seondary hydrogen = 3.9 X rate of abstraction of primary hydrogen
probability of formation of isopropyl chloride = 3.9 X 1 (relative rate X relative number of secondary hydrogens)
Probability of formation of n-propyl chloride = 1 X 3 (relative rate X relative number of primary hydrogens)
Total probability = 3.9
% of n-propyl chloride = 3 X 100 / 6.9 = 43.48 %
They will be stay the same all the way through. your product must be the same mass
