What atomic or hybrid orbitals make up the sigma bond between Cl and F in chlorine trifluoride, ClF3
1 answer:
Answer:
sp³d¹ hybridization
Explanation:
Given Cl as central element with three F substrates ...
The VSEPR structure indicates 5 hybrid orbitals that contain 2 diamagnetic orbitals (non-bonded e⁻-pairs) and 3 paramagnetic orbitals (single, non-paired electron for covalent bonding with fluorine) giving a trigonal bypyrimidal parent with a T-shaped geometry.
Valence bond theory predicts the following during bonding:
Cl:[Ne]3s²3p²p²p¹3d⁰
=> [Ne]3s²p²p¹p¹d¹
=> [Ne]3(sp³d)²(sp³d)²(sp³d)¹(sp³d)¹(sp³d)¹
giving 3 ( [Cl](sp³d) - [F]2p¹ ) sigma bonds and 2 non-bonded pairs on Cl.
Note the following images:
Non-bonded electron pairs are in plane of parent geometry and Fluorides covalently bonded to central element Chloride forming the T-shaped geometry.
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Answer:
Explanation:
The unbalanced nuclear equation is
It is convenient to replace the question by an atomic symbol, , where <em>x </em>= the atomic number, <em>y</em> = the mass number, and Z = the symbol of the element.
Then your equation becomes
The main point to remember in balancing nuclear equations is that the <em>sums of the superscripts and of the subscripts</em> must be the same on each side of the equation.
Then
93 – 1 = <em>x</em>, so <em>x</em> = 92
232 + 0 = <em>y</em>, so <em>y</em> = 232
Element 92 is uranium, so the nuclear equation becomes