Answer:
5 electron groups, see saw
Explanation:
During the formation of SF4, the sulfur atom usually bonds with each of four fluorine atoms where 8 of valence electrons are used. The four fluorine atoms have 3 lone pairs of electrons in its octet which will further utilize 24 valence electrons. In addition, two electrons are present as a lone pair on the sulfur atom. We can determine sulfur’s hybridization state by counting of the number of regions of electron density on sulphur (the central atom in the molecule). When bonding takes place there is a formation of 4 single bonds to sulfur and it has 1 lone pair. Looking at this, we can say that the number of regions of electron density is 5. The hybridization state is sp3d.
SF4 molecular geometry is seesaw with one pair of valence electrons. The molecule is polar. The equatorial fluorine atoms have 102° bond angles instead of the actual 120° angle. The axial fluorine atom angle is 173° instead of the actual 180° bond angle.
Answer:
3 : 1
Explanation:
Let the rate of He be R1
Molar Mass of He (M1) = 4g/mol
Let the rate of O2 be R2
Molar Mass of O2 (M2) = 32g/mol
Recall:
R1/R2 = √(M2/M1)
R1/R2 = √(32/4)
R1/R2 = √8
R1/R2 = 3
The ratio of rate of effusion of Helium to oxygen is 3 : 1
Explanation:
When two small nuclei combine together to form a large nuclei then it is known as nuclear fusion.
When nuclei of two hydrogen atoms fuse together then it results in the formation of a helium atom along with the release of lot of energy. This energy is nuclear energy.
This nuclear reaction is as follows.
Thus, we can conclude that nuclear fusion represents nuclear energy.
B is the correct answer to your question
