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m_a_m_a [10]
4 years ago
12

In the molecule ClF3, chlorine makes three covalent bonds. Therefore, three of its seven valence electrons need to be unpaired.

The orbitals with the same energy are known as degenerate orbitals. For example, the p subshell has three degenerate orbital, namely, px, py, and pz. How many degenerate orbitals are needed to contain seven electrons with three of them unpaired
Chemistry
1 answer:
Kisachek [45]4 years ago
3 0

Answer:

Explanation:

Chlorine has electronic configuration of 2 , 8 , 7

In n = 3 there are 7 electrons out of which 2 are in s , and 5 are in p . But out of 5 electrons in p , one electron jumps into d orbital . so the electronic configuration  becomes as follows

3s^23p_x^23p_y^23p_z^1  = 7

3s^23p_x^23p_y^13p_z^13d_{xy}^1

These orbitals like sp³d hybridise to form 7 degenerate orbitals out of which 2 orbitals contain electrons in pairs and rest three are singly occupied by electrons.( unpaired electrons )

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9.92*10^{-5}molNH_{3}*\frac{3molOBr^{-} }{2molNH_{3}}=   1.488*10^{-4}mol OBr^{-}

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Molarity of hypobromite solution = \frac{1.488*10^{-4}molOBr^{-}}{1.00mL} *\frac{1000mL}{1L} =0.1488mol/L

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