Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Use data in this chapter and Chapter 12 to discuss why MgO is an ionic compound but CO is not an ionic compound.

Short Answer

Expert verified

Due to significant difference between electronegativity of Mg and O, they form an ionic bond. But in CO, the electronegativity difference between C and O is not very large. Thus they form covalent bond by mutual sharing of electrons.

Step by step solution

01

- Determining the MgO:

In the case of the MgO compound, magnesium is a metal atom that belongs to the alkali group and oxygen is a nonmetal that belongs to the p-block group. So, magnesium metal transfers its electrons whereas oxygen atoms will accept it and a bond will form between them which is called an ionic bond.

02

- Determining that how magnesium metal is transferring its electron:

The magnesium metal is transferring its electrons and oxygen as a nonmetal accepting the electrons, and both these formed Mg+2and O-2ions which are held together by the electrostatic force of attraction, and an ionic bond forms between them.

03

- Determine the CO compound:

CO is a covalent compound, as firstly, it consists of a double covalent bond and a single coordinate covalent bond (in total, 3 covalent bonds), between carbon and oxygen atoms. Secondly, due to a small difference of electronegativities of carbon and oxygen atoms (∆E=1.0), these atoms settle upon sharing of electrons.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

All amino acids have at least two functional groups with acidic or basic properties. In alanine the carboxylic acid group has Ka= 4.5×10- 3and the amino group has Kb= 7.4×10- 5. Three ions of alanine are possible when alanine is dissolved in water. Which of these ions would predominate in a solution with [H+]= 1.0Min a solution with [OH-]= 1.0M?

Consider the following reaction:

CH3X + Y → CH3Y + X

At 25·C the following two experiments were run, yielding the following data:

Experiment 1: [Y]0 = 3.0 M

Experiment 2: [Y]0 = 4.5 M

Experiments were also run at 85·C. The value of the rate constant at 85·C was found to be 7.88 × 108 (with the

time in units of hours), where [CH3X]0 = 1.0 × 10-2 Mand [Y]0 = 3.0 M.

a. Determine the rate law and the value of kfor this reaction at 25·C.

b. Determine the half-life at 85·C

c. Determine Eafor the reaction.

d. Given that the C-X bond energy is known to be about 325 kJ/mol, suggest a mechanism that explains the results in parts a and c.

If the fluid inside a tree is about 0.1M more concentrated in solute than the groundwater that bathes the roots, how high will the column of fluid rise in the tree at 250C? Assume that the density of the fluid is 1.0 g/cm3. The density of mercury is 13.6g/cm3

The rate law for the decomposition of phosphine (PH3) is Rate=d[PH3]/dt=k[PH3]. It takes 120s for the concentration of 1.00 M PH3 to decrease to 0.250M. How much time is required for 2.00 M(PH3) to decrease to a concentration of 0.350 M?

Question: Consider a sample containing 2.00 moles of a monatomic ideal gas that undergoes the following changes:

For each step, assume that the external pressure is constant and equals the final pressure of the gas for that step. Calculate q, w, ΔE, and ΔH for each step and for the overall change from state A to state D

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free