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2 years ago

Which one of the following molecules has an atom with an incomplete octet? A. BF3 B. GeH4 C. NF3 D. AsCl3 E. H2O

Chemistry

1 answer:

Murrr4er [49]2 years ago

7 0

Answer:

$BF_{3}$ has an atom with incomplete octet

Explanation:

H atoms are generally excluded from octet rule as H atom has only one 1s orbital which can contain maximum two electrons.

At first, lewis structures of all molecules should be drawn. Then octet of an atom is determined by taking summation of all nonbonding electrons along with all bonding electrons corresponding to that atom.

If this summation is less than 8 then it is said that the atom has incomplete octet.

In $BF_{3}$ , B atom has no nonbonding electons (i.e. lone pair). B atom is attached to three F atoms via three single bonds. Then total number of electrons corresponding to B atom are 6 ( one bond contains two electrons).

All other atoms (except H) in given molecules has complete octet.

So, $BF_{3}$ has an atom with incomplete octet.

Lewis structures are shown below.

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Sodium reacts with chlorine gas according to the following reaction: 2Na(s)+Cl2(g)→2NaCl(s) What volume of Cl2 gas, measured at

asambeis [7]

Answer:6.719Litres of Cl2 gas.

Explanation:According to eqn of rxn

2Na +Cl2=2NaCl

P=689torr=689/760=0.91atm

T=39°C+273=312K

according to stoichiometry of the reaction,1Moles of Cl2 gives 2moles of NaCl

But 28g of NaCl was given,we have to convert this to moles by using the relation, n=mass/MW

MW of NaCl=23+35.5=58.5g/mol

n=28g(mass given of NaCl)/58.5

n=0.479moles of NaCl

Going back to the reaction,

if 1moles of Cl2 produces 2moles of NaCl

x moles of Cl2 will give 0.479moles of NaCl.

x=0.479*1/2

x=0.239moles of Cl2.

To find the volume, we use ideal ggas eqn,PV=nRT

V=nRT/P

V=0.239*0.082*312/0.91

V=6.719Litres

6 0

2 years ago

A person loses 2.70 lbs in two weeks how many grams did he lose?

Evgesh-ka [11]

He person lost 1224.7 grams.

6 0

3 years ago

Read 2 more answers

A solution is prepared by mixing 50.0 mL of 0.50 M Cu(NO3)2 with 50.0 mL of 0.50 M Co(NO3)2. Sodium hydroxide is added to the mi

vekshin1

Answer:

Cu(OH)₂ will precipitate first, with [OH⁻] = 2.97x10⁻¹⁰ M

Explanation:

The equilibriums that take place are:

Cu⁺² + 2OH⁻ ↔ Cu(OH)₂(s) ksp = 2.2x10⁻²⁰ = [Cu⁺²]*[OH⁻]²

Co⁺² + 2OH⁻ ↔ Co(OH)₂(s) ksp = 1.3x10⁻¹⁵ = [Co⁺²]*[OH⁻]²

Keep in mind that <em>the concentration of each ion is halved </em>because of the dilution when mixing the solutions.

For Cu⁺²:

2.2x10⁻²⁰ = [Cu⁺²]*[OH⁻]²

2.2x10⁻²⁰ = 0.25 M*[OH⁻]²

[OH⁻] = 2.97x10⁻¹⁰ M

For Co⁺²:

1.3x10⁻¹⁵ = [Co⁺²]*[OH⁻]²

1.3x10⁻¹⁵ = 0.25 M*[OH⁻]²

[OH⁻] = 7.21x10⁻⁸ M

<u>Because Copper requires less concentration of OH⁻ than Cobalt</u>, Cu(OH)₂ will precipitate first, with [OH⁻] = 2.97x10⁻¹⁰ M

3 0

2 years ago

Which of the following type of protons are chemically equivalent? A) homotopic B) enantiotopic C) diastereotopic A & B B &am

Rina8888 [55]

Answer:

A) homotopic and B) enantiotopic

Explanation:

Protons chemically equivalent are those that have the same chemical shift, also if they are interchangeable by some symmetry operation or by a rapid chemical process.

The existence of symmetry axes, Cn, that relate to the protons results in the protons being homotopic, that is chemically equivalent in both chiral and aquiral environments.

The existence of a plane of symmetry, σ, makes the protons related by it, are enantiotopic and these protons will only be equivalent in an aquiral medium; if the medium is chiral both protons will be chemically NOT equivalent. The existence of a center of symmetry, i, in the molecule makes the related protons through it enantiotopic and therefore chemically only in the aquiral medium.

Diastereotopic protons cannot be interconverted by any symmetry operation and they are different, with different chemical displacement.

6 0

2 years ago

Sulfurous acid is a diprotic acid with the following acid-ionization constants: Ka1 = 1.4x10−2, Ka2 = 6.5x10−8 If you have a 1.0

REY [17]

Answer:

pH = 7.1581

Explanation:

The equilibrium of NaHSO₃ with Na₂SO₃ is:

HSO₃⁻ ⇄ SO₃²⁻ + H⁺

<em>Where K of equilibrium is the Ka2: 6.5x10⁻⁸</em>

HSO₃⁺ reacts with NaOH, thus:

HSO₃⁻ + NaOH → SO₃²⁻ + H₂O + Na⁺

As the buffer is of 1.0L, initial moles of HSO₃⁻ and SO₃²⁻ are:

HSO₃⁻: 0.252 moles

SO₃²⁻: 0.139 moles

Based on the reaction of NaOH, moles added of NaOH are subtracting moles of HSO₃⁻ and producing SO₃²⁻. The moles added are:

0.0500L ₓ (1mol /L): 0.050 moles of NaOH.

Thus, final moles of both compounds are:

HSO₃⁻: 0.252 moles - 0.050 moles = 0.202 moles

SO₃²⁻: 0.139 moles + 0.050 moles = 0.189 moles

Using H-H equation for the HSO₃⁻ // SO₃²⁻ buffer:

pH = pka + log [SO₃²⁻] / [HSO₃⁻]

Where pKa is - log Ka = 7.187

Replacing:

pH = 7.187 + log [0.189] / [0.202]

4 0

2 years ago