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kolezko [41]
3 years ago
7

In addition to NF3, two other fluoro derivatives of nitrogen are known: N2F4 and N2F2. What shapes do you predict for these two

molecules?

Chemistry
1 answer:
sineoko [7]3 years ago
7 0

Answer:

Pyramid trigonal and trigonal planar, respectively.

Explanation:

The shape of a molecule is how the atoms are organized in the space, and it happens to minimize the repulsive force of the bonds and the lone pairs of electrons. Thus, in the molecules given, the two N atoms are the central atoms, because they can do the most number of bonds.

Nitrogen has 5 electrons in the valence shell, so it can do 3 bonds to be stable with 8 (octet rule), and fluorine has 7 electrons in the electron shell, so it can do 1 bond to be stable with 8.

In the molecule of N2F4, the two nitrogen do a simple bond between then, and simple bond with 2 F each, as shown below. So, each nitrogen still has 1 lone pair of electrons. To minimize it, the better shape is the pyramid trigonal.

In the molecule of N2F2, the two nitrogen do a double bond between them, and a simple bond with one F each, as shown below. They still have lone pairs, and the double bond is stiff, so it doesn't rotate. Thus, the trigonal planar shape is the better one.

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How many unpaired electrons are in the c 2 + ​ ion?
sladkih [1.3K]
Atomic number of C is 6. Hence, there are 6 electrons in carbon.

The electronic configuration of carbon is 1s2, 2s2, 2p2

Here, there are 2 unpaired electron. However, C2+ ions have 2 electrons less as compared to C.

Hence, electronic configuration of C 2+ ion is 1s2, 2s2. All the electrons are paired in this system. So there are no unpaired electrons in C 2+ ion. 
7 0
3 years ago
Help pls again HAHAHAH
Alenkasestr [34]

<u><em>I believe the answer you are looking for is position 4 </em></u>, because the northern face of the hemisphere is facing away from the sun not getting

to much heat nor daylight therefore its cold making it winter fully in option 4.

Just a tip option 3 looks like its facing back but half of it is still shown to the sun.

5 0
3 years ago
Consider the following ionic compounds: CdCO3, Na2S, PbSO4, (NH4)3PO4, and Hg2Cl2. Which compounds will be soluble when added to
k0ka [10]

Answer:

Na2S, (NH4)3PO4

Explanation:

We can decide what compounds are soluble by considering the solubility rules that apply.

CdCO3 is not soluble in water because all carbonates are insoluble except those of ammonium, sodium and potassium.

Na2S is soluble in water because all sulphides are insoluble except those of sodium, potassium and ammonium.

PbSO4 is insoluble in water because all sulphates are soluble except those of lead and barium. The sulphate of calcium is only slightly soluble in water.

(NH4)3PO4 is soluble in water because all phosphates are insoluble except those of sodium, potassium and ammonium.

Hg2Cl2 is insoluble in water because all chlorides are soluble except those of lead, mercury II and silver.

7 0
3 years ago
Calculate molality,molarity and mole fraction of KI if the density of 20%(mass) aqueous KI is 1.202 g/mL
Scorpion4ik [409]
Basis: 1 L of the substance.
               (1.202 g/mL) x (1000 mL) = 1202 g 
                   mass solute = (1202 g) x 0.2 = 240.2 g
                   mass solvent = 1202 g x 0.8 =  961.6 g
                   moles KI = (240.2 g) x (1 mole / 166 g)  = 1.45 moles
                   moles water = (961.6 g) x (1 mole / 18 g) = 53.42 moles
1. Molality = moles solute / kg solvent 
                  = 1.45 moles / 0.9616 kg = 1.5 m
2. Molarity = moles solute / L solution
                  = 1.45 moles / 1 L solution = 1.45 M
3. molar mass = mole solute / total moles
                        =  1.45 moles / (1.45 moles + 53.42 moles) = 0.0264 

5 0
3 years ago
Why is a terminal alkyne favored when sodium amide (NaNH2) is used in an elimination reaction with 2,3-dichlorohexane
mihalych1998 [28]

The question is incomplete, the complete question is;

Why is a terminal alkyne favored when sodium amide (NaNH2) is used in an elimination reaction with 2,3-dichlorohexane? product. A) The terminal alkyne is more stable than the internal alkyne and is naturally the favored B) The terminal alkyne is not favored in this reaction. C) The resonance favors the formation of the terminal rather than internal alkyne. D) The strong base deprotonates the terminal alkyne and removes it from the equilibrium.

E) The positions of the Cl atoms induce the net formation of the terminal alkyne.

Answer:

E) The positions of the Cl atoms induce the net formation of the terminal alkyne.

Explanation:

In this reaction, sterric hindrance plays a very important role. We know that sodamide is a strong base, it tends to attack at the most accessible position.

The first deprotonation yields an alkene. The strong base attacks at the terminal position again and yields the terminal alkyne. Thus the structure of the dihalide makes the terminal hydrogen atoms most accessible to the base. Hence the answer.

7 0
3 years ago
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