According to the molecular geometry, there are two lone pairs on central atom of BrF₃.
Molecular geometry can be defined as a three -dimensional arrangement of atoms which constitute the molecule.It includes parameters like bond length,bond angle and torsional angles.
It influences many properties of molecules like reactivity,polarity color,magnetism .The molecular geometry can be determined by various spectroscopic methods and diffraction methods , some of which are infrared,microwave and Raman spectroscopy.
They provide information about geometry by taking into considerations the vibrational and rotational absorbance of a substance.Neutron and electron diffraction techniques provide information about the distance between nuclei and electron density.
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Answer:
BrF3 has 2 lone pairs (4 unshared electrons)
Explanation:
Answer: Cis-1-bromo-4-tert-butylcyclohexane would undergo faster elimination reaction.
Explanation:
The two primary requirements for an E-2 elimination reaction are:
1.There must be availability of β-hydrogens that is presence of hydrogen on the carbon next to the leaving group.
2.The hydrogen and leaving group must have a anti-periplanar position .
Any substrate which would follow the above two requirements can give elimination reactions.
For the structure of trans-1-bromo-4-tert-butylcyclohexane and cis-1-bromo-4-tert-butylcyclohexane to be stable it must have the tert-butyl group in the equatorial position as it is a bulky group and at equatorial position it would not repel other groups. If it is kept on the axial position it would undergo 1,3-diaxial interaction and would destabilize the system and that structure would be unstable.
Kindly find the structures of trans-1-bromo-4-tert-butylcyclohexane and cis-1-bromo-4-tert-butylcyclohexane in attachment.
The cis- 1-bromo-4-tert-butylcyclohexane has the leaving group and β hydrogens in anti-periplanar position so they can give the E2 elimination reactions easily.
The trans-1-bromo-4-tert-butylcyclohexane does not have the leaving group and βhydrogen in anti periplanar position so they would not give elimination reaction easily.
so only the cis-1-bromo-4-tert butyl cyclohexane would give elimination reaction.
Trans-1-bromo-4-tert-butylcyclohexane is expected to undergo E2 elimination faster than cis-1-bromo-4-tert-butylcyclohexane due to less steric hindrance.
In determining the rate of E2 elimination, the trans-1-bromo-4-tert-butylcyclohexane would undergo E2 elimination faster than the cis-1-bromo-4-tert-butylcyclohexane. This is due to the larger degree of steric hindrance in the case of the cis isomer.
In trans-1-bromo-4-tert-butylcyclohexane, the bromine is at the equatorial position while the tert-butyl group is axial. It forms a structure that allows the compound to experience less steric hindrance with bromine in a more favorable position for leaving.
In comparison, cis-1-bromo-4-tert-butylcyclohexane has a bromine and tert-butyl group both at equatorial positions. This causes steric hindrance, and in turn, slows down the E2 elimination rate. Despite the more stable conformation, the bromine is not well-oriented for a leaving group in E2 elimination.
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1s22s22p3
1s22s22p4
1s22s22p5
The electronic configuration that matches that of the model is 1s²2s²2p³; option B.
Electronic configuration of atoms of elements is the arrangement of electrons in the electron shells or orbitals in atom.
The electronic configuration of an atom depends on the atomic number of the element.
The atomic number of nitrogen is 7. This means that it has seven electrons in its orbitals.
The electronic configuration that matches that of the nitrogen atom is 1s²2s²2p³.
In conclusion, the electronic configuration of nitrogen indicates that it has seven electrons in its orbitals.
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Note that the complete question is given below:
Consider the model of the nitrogen atom. Which electron configuration matches this model?
1s22s22p2
1s22s22p3
1s22s22p4
1s22s22p5
molecular formula of tetraethyl pyrophosphate is C8H20O7P2
The molecular formula for TEPP, or tetraethyl pyrophosphate, is approximately C4H10O6P2. This is calculated from the given element percentages and the given molecular mass.
The process of figuring out the molecular formula of TEPP, which stands for tetraethyl pyrophosphate, involves several steps. First, given the percentage composition of each element, we need to calculate the number of moles of each element in a 100g sample of the compound. For example, the 33.11% carbon equals 33.11g in a 100g sample. When divided by the molar mass of carbon (12.01 g/mole), this gives us approximately 2.76 moles. We do the same calculation for hydrogen, oxygen, and phosphorus to end up with a 'base' empirical formula of C2.76H6.96O3.86P1.36.
The next step is to figure out how to get the simplest whole number ratio. This can be done by dividing all the numbers by the smallest one, 1.36. This gives us C2.03H5.12O2.84: these are approximately 2, 5, 3, and 1, respectively, so the empirical formula of TEPP is approximately C2H5O3P.
We need to compare the empirical mass to the given molecular mass of 290.19 g/mol to determine whether the molecular formula is a multiple of the empirical. The empirical formula mass is about 144.99 g/mol. Therefore, the molecular formula of TEPP is approximately C4H10O6P2, as the molecular weight is approximately twice that of the empirical formula weight.
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Answer:
More reactants will be produced
Explanation:
Le Chatelier's principle; adding additional product or reactant will move the equilibrium left or right to compensate and come back to equilibrium
By adding more product to your system at equilibrium, the equilibrium will shift towards reactants, more reactants will be produced
Answer: Liquid molecules forming a gas and gas molecules forming a liquid are equal in number
Explanation: :/
A. Gain 2 electrons
B. Gain 3 electrons
C. Lose 2 electrons
D. Lose 3 electrons
Nitrogen changes from +5 in - to +2 in NO. This means nitrogen has gained 3 electrons. Option B
To determine the change in electrons for nitrogen in the given reaction, we need to compare the oxidation state of nitrogen in the reactant (-) and the product (NO).
In the reactant, -, nitrogen is in the +5 oxidation state. This is because oxygen has an oxidation state of -2, and there are three oxygen atoms in -. Therefore, nitrogen must have an oxidation state of +5 to balance the overall charge of -.
In the product, NO, nitrogen is in the +2 oxidation state. This is because oxygen has an oxidation state of -2, and there is only one oxygen atom in NO. Therefore, nitrogen must have an oxidation state of +2 to balance the overall charge of NO.
By comparing the oxidation states of nitrogen in the reactant and the product, we can determine the change in electrons. The change in oxidation state corresponds to the change in the number of electrons gained or lost by the nitrogen atom.
In this case, nitrogen changes from +5 in - to +2 in NO. This means nitrogen has gained 3 electrons.
Therefore, the correct answer is B) Gain 3 electrons.
The nitrogen atom undergoes a reduction because it gains electrons, reducing its oxidation state from +5 to +2 in the reaction.
Option B
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