The correct answer is C. square planar
According to the Valence Shell Electron Pair Repulsion Theory(VSEPR), The shape of a molecule depends on the number of electron pairs in the molecule.
VSEPR theory was first coined by Gillespie and Nyhlom in 1957 as an improvement over the Sidgwick - Powell theory.
According to this theory, the shape of a molecule is determined by the number of electron pairs that surround the valence shell of the central atom in the molecule. The electron pairs are positioned as far apart in space as possible to minimize repulsion of electron pairs.
However, the presence of lone pairs distorts the shape anticipated for the molecule on the basis of VSEPR.
For a molecule having six electron pairs, an octahedral geometry is expected(electron domain geometry). However, the presence of two lone pairs which are positioned at opposite side of the four single bonds leads to an observed square planar molecular geometry.
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Answer:
D. The cation will use a type II naming system
Explanation:
As the element becomes stable, Option D is NOT taking place. That means that as the element becomes stable, the cation will NOT use a type II naming system. This is because in Type II binary compounds, the metal is not found in either Group 1 or Group 2 on the periodic table. It is as a result of this that the cation will not use a type II naming system.
Group 1 and Group 2 elements make use of type I naming system because in Type I binary compounds, the metal is present.
Option D is the correct answer.
So 575 ft=175260mm so all you have to do is divide
175260/1.55=<span>113070.967742</span>
Hybrid Orbitals: are used to describe the orbitals in covalently bonded atoms sp,sp2,sp3
You can find the hybridization by adding the number of bonded atoms and the number of lone pairs.
For example in BF3. The central atom (B) is bonded to three atoms. So the hybridization is sp2
In NH3, the central atom (N) is bonded to three atoms and has one lone pair. The hybridization is sp3
The empirical formula for this would be FeS
